3.Hydrogen Gas Burns in Air According to the Equation Below

Problems 2 & 3 – ANSWERS

2011 A

3.Hydrogen gas burns in air according to the equation below.

2 H2(g) + O2(g) 2 H2O(l)

(a)Calculate the standard enthalpy change, ∆H°298, for the reaction represented by the equation above. (The molar enthalpy of formation, ∆Hf°, for H2O(l) is -285.8 kJ mol-1 at 298 K.)

(b)Calculate the amount of heat, in kJ, that is released when 10.0 g of H2(g) is burned in air.

(c)Given that the molar enthalpy of vaporization, ∆H°vap, for H2O(l) is 44.0 kJ mol-1 at 298 K, what is the standard enthalpy change, ∆H°298 for the reaction 2 H2(g) + O2(g) 2 H2O(g) ?

A fuel cell is an electrochemical cell that converts the chemical energy stored in a fuel into electrical energy. A cell that uses H2 as the fuel can be constructed based on the following half-reactions.

Half-reaction / E° (298 K)
2 H2O(l) + O2(g) + 4 e– 4 OH–(aq) / 0.40 V
2 H2O(l) + 2 e– H2(g) + 2 OH–(aq) / –0.83 V

(d)Write the equation for the overall cell reaction.

(e)Calculate the standard potential for the cell at 298 K.

(f)Assume that 0.93 mol of H2(g) is consumed as the cell operates for 600. seconds.

(i)Calculate the number of moles of electrons that pass through the cell.

(ii)Calculate the average current, in amperes, that passes through the cell.

(g)Some fuel cells use butane gas, C4H10 rather than hydrogen gas. The overall reaction that occurs in a butane fuel cell is 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l). What is one environmental advantage of using fuel cells that are based on hydrogen rather than on hydrocarbons such as butane?

Answer:

(a)∆Hrxn = ∆Hprod – ∆Hreact = (2)(–285.8 kJ mol-1) – 0 = –571.6 kJ

(b)= 1420 kJ

2 H2O(l) 2 H2O(g)∆H = (2)(+44.0 kJ mol-1)

2 H2(g) + O2(g) 2 H2O(g)∆H = -483.6 kJ

(d)2 H2O(l) + O2(g) + 4 e– 4 OH–(aq)E° = +0.40 V

2 H2(g) + 4 OH–(aq) 4 H2O(l) + 4 e–E° = +0.83 V

2 H2(g) + O2(g) 4 H2O

(e)0.40 V + 0.83 V = 1.23 V

(f)(i) 0.93 mol H2= 1.86 mol e–

(ii) amp  sec. = coul; amp = = 3.0102 amp

(g)hydrogen gas fuel cells do not add CO2, a greenhouse gas, to the environment

2010 A

8 H+(aq) + 4 Cl-(aq) + MnO4-(aq) 2 Cl2(g) + Mn3+(aq) + 4 H2O(l)

3.Cl2(g) can be generated in the laboratory by reacting potassium permanganate with an acidified solution of sodium chloride. The net-ionic equation for the reaction is given above.

(a)A 25.00 mL sample of 0.250 M NaCl reacts completely with excess KMnO4(aq). The Cl2(g) produced is dried and stored in a sealed container. At 22°C the pressure of the Cl2(g) in the container is 0.950 atm.

(i)Calculate the number of moles of Cl-(aq) present before any reaction occurs.

(ii)Calculate the volume, in L, of the Cl2(g) in the sealed container.

An initial-rate study was performed on the reaction system. Data for the experiment are given in the table below.

Trial / [Cl-] / [MnO4-] / [H+] / Rate of Disappearance of MnO4- in M s-1
1 / 0.0104 / 0.00400 / 3.00 / 2.25  10-8
2 / 0.0312 / 0.00400 / 3.00 / 2.03  10-7
3 / 0.0312 / 0.00200 / 3.00 / 1.02  10-7

(b)Using the information in the table, determine the order of the reaction with respect to each of the following. Justify your answers.

(i)Cl-

(ii)MnO4-

(c) The reaction is known to be third order with respect to H+. Using this information and your answers to part (b) above, complete both of the following:

(i) Write the rate law for the reaction.

(ii) Calculate the value of the rate constant, k, for the reaction, including appropriate units.

(d) Is it likely that the reaction occurs in a single elementary step? Justify your answer.

Answer:

(a)(i) 25.00 mL = 0.00625 mol Cl-

(ii) 0.00625 mol Cl-= 0.00313 mol Cl2

V = = 0.0796694 = 0.0797 L

(b)rate = k[Cl-]a[MnO4-]b[H+]c ; = k[H+]c ; k[H+]c is a constant for all trials

(i) comparing trial 1 with trial 2 and eliminating [MnO4-] since they are equal, we get ; solving results in a = 2, therefore, 2nd order with respect to [Cl-]

(ii) comparing trial 2 with trial 3 and eliminating [Cl-] since they are equal, we get ; solving results in b = 1, therefore, 1st order with respect to [MnO4-].

(c)(i) rate = k[Cl-]2[MnO4-]1[H+]3

(ii) trial 1; 2.25  10-8M s-1 = k(0.0104 M)2(0.00400 M)1(3.00 M)3 ; k = 0.00193 M-5 s-1

(d)it is NOT likely to occur in a single elementary step; in order for that to occur over 3 particles would need to collide in the same spot with the proper orientation and energy. This very highly unlikely event would not be expected to happen, near zero probability.

2009 A

CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g)

3.Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by the equation above.

(a)A 25.0 g sample of methane is placed in a reaction vessel containing 2.58 mol of Cl2(g).

(i)Identify the limiting reactant when the methane and chlorine gases are combined. Justify your answer with a calculation.

(ii)Calculate the total number of moles of CH2Cl2(g) in the container after the limiting reactant has been totally consumed.

Initiating most reactions involving chlorine gas involves breaking the Cl-Cl bond, which has a bond energy of 242 kJ mol-1.

(b)Calculate the amount of energy, in joules, needed to break a single Cl-Cl bond.

(c)Calculate the longest wavelength of light, in meters, that can supply the energy per photon necessary to break the Cl-Cl bond.

The following mechanism has been proposed for the reaction of methane gas with chlorine gas. All species are in the gas phase.

Step 1Cl2 2 Clfast equilibrium

Step 2CH4 + Cl CH3 + HClslow

Step 3CH3 + Cl2 CH3Cl + Clfast

Step 4CH3Cl + Cl CH2Cl2 + Hfast

Step 5H + Cl HClfast

(d)In the mechanism, is CH3Cl a catalyst, or is it an intermediate? Justify your answer.

(e)Identify the order of the reaction with respect to each of the following according to the mechanism. In each case, justify your answer.

(i)CH4(g)

(ii)Cl2(g)

Answer:

(a)(i) 25.0 g CH4= 1.56 mol CH2Cl2

2.58 mol Cl2= 1.29 mol CH2Cl2

Cl2 is the limiting reactant

(ii) 1.29 mol CH2Cl2 when you run out of chlorine

(b)1 Cl-Cl bond = 4.0210-19 J

(c)= 4.9510-7 m

(d)intermediate; it is made in step 3 and consumed in step 4, to be a catalyst it would have to remain unchanged at the end

(e)the slowest step (step 2) is the rate determining step

rate of step 1: ratefor = kfor[Cl2] , raterev = krev[Cl]2

at equilibrium: ratefor = raterev, kfor[Cl2] = krev[Cl]2

[Cl] =[Cl2]1/2 = k1[Cl2]1/2

rate of step 2: rate = k2[CH4][Cl] ; substituting, rate = k[CH4][Cl2]1/2

therefore, (i) rate is 1st order with respect to CH4 and (ii) ½ order with respect to Cl2

2006 B

CO(g) + O2(g) CO2(g)

The combustion of carbon monoxide is represented by the equation above.

(a)Determine the value of the standard enthalpy change, ∆H˚rxn for the combustion of CO(g) at 298 K using the following information.

C(s) + O2(g) CO(g)∆H˚298 = –110.5 kJ mol-1

C(s) + O2(g) CO2(g)∆H˚298 = –393.5 kJ mol-1

(b)Determine the value of the standard entropy change, ∆S˚rxn, for the combustion of CO(g) at 298 K using the information in the following table.

Substance / S˚298
(J mol-1 K-1)
CO(g) / 197.7
CO2(g) / 213.7
O2(g) / 205.1

(c)Determine the standard free energy change, ∆G˚rxn, for the reaction at 298 K. Include units with your answer.

(d)Is the reaction spontaneous under standard conditions at 298 K? Justify your answer.

(e)Calculate the value of the equilibrium constant, Keq, for the reaction at 298 K.

Answer:

(a)∆H˚rxn == (–393.5 kJ mol-1) – (–110.5 kJ mol-1 + 1/2(0)) = –283.0 kJ

(b)∆S˚rxn == (213.7) – (197.7 + 1/2(205.1) J mol-1 K-1 = -86.6 J K-1

(c)∆G˚ = ∆H˚ – T∆S˚ = (–283.0 kJ) – (298K)(-0.0866 J K-1) = –257.2 kJ

(d)spontaneous; any reaction in which the ∆G˚ < 0 is spontaneous

(e)Keq = e–(∆G/RT) = e–(–257208.1J/(8.31298) = 1.28  1045

2004 B

2 Fe(s) + O2(g) Fe2O3(s)∆Hf˚ = -824 kJ mol–1

Iron reacts with oxygen to produce iron(III) oxide as represented above. A 75.0 g sample of Fe(s) is mixed with 11.5 L of O2(g) at 2.66 atm and 298 K.

(a)Calculate the number of moles of each of the following before the reaction occurs.

(i)Fe(s)

(ii)O2(g)

(b)Identify the limiting reactant when the mixture is heated to produce Fe2O3. Support your answer with calculations.

(c)Calculate the number of moles of Fe2O3 produced when the reaction proceeds to completion.

(d)The standard free energy of formation, ∆Gf˚ of Fe2O3 is –740. kJ mol–1 at 298 K.

(i)Calculate the standard entropy of formation ∆Sf˚ of Fe2O3 at 298 K. Include units with your answer.

(ii)Which is more responsible for the spontaneity of the formation reaction at 298K, the standard enthalpy or the standard entropy?

The reaction represented below also produces iron(III) oxide. The value of ∆H˚ for the reaction is –280 kJ per mol.

2 FeO(s) + O2(g) Fe2O3(s)

(e)Calculate the standard enthalpy of formation, ∆Hf˚ of FeO(s).

Answer:

(a)(i) 75.0 g Fe = 1.34 mol Fe

(ii) PV = nRT, n =

= 1.25 mol O2

(b)Fe; 1.34 mol Fe = 1.01 mol O2

excess O2, limiting reagent is Fe

(d)(i) ∆Gf˚ = ∆Hf˚ – T∆Sf˚

–740 kJ mol–1 = –824 kJ mol–1 – (298 K)(∆Sf˚)

∆Sf˚ = 0.282 kJ mol–1 K–1

(ii) standard enthalpy; entropy decreases (a non-spontaneous process) so a large change in enthalpy (exothermic) is need to make this reaction spontaneous

(e)∆H = ∆Hf(products) – ∆Hf(reactants)

–280 kJ mol–1 = –824 kJ mol–1 – [2(∆Hf˚ FeO) – 1/2(0)]

= -272 kJ mol–1

2005 B

Answer the following questions related to the kinetics of chemical reactions.

I–(aq) + ClO–(aq) IO–(aq) + Cl–(aq)

Iodide ion, I–, is oxidized to hypoiodite ion, IO–, by hypochlorite, ClO–, in basic solution according to the equation above. Three initial-rate experiments were conducted; the results shown in the following table.

Experiment / [I–]
(mol L–1) / [ClO–]
(mol L–1) / Initial Rate of Formation of IO– (mol L–1 s–1)
1 / 0.017 / 0.015 / 0.156
2 / 0.052 / 0.015 / 0.476
3 / 0.016 / 0.061 / 0.596

(a)Determine the order of the reaction with respect to each reactant listed below. Show your work.

(i)I–(aq)

(ii)ClO–(aq)

(b)For the reaction,

(i)write the rate law that is consistent with the calculations in part (a);

(ii)calculate the value of the specific rate constant, k, and specify units.

The catalyzed decomposition of hydrogen peroxide, H2O2(aq), is represented by the following equation.

2 H2O2(aq) 2 H2O(l) + O2(g)

The kinetics of the decomposition reaction were studied and the analysis of the results show that it is a first-order reaction. Some of the experimental data are shown in the table below.

[H2O2]
(mol L–1) / Time (minutes)
1.00 / 0.0
0.78 / 5.0
0.61 / 10.0

(c)During the analysis of the data, the graph below was produced.

(i)Label the vertical axis of the graph

(ii)What are the units of the rate constant, k, for the decomposition of H2O2(aq) ?

(iii)On the graph, draw the line that represents the plot of the uncatalyzed first-order decomposition of 1.00 M H2O2(aq).

Answer:

(a)(i) comparing expt. 1 to expt. 2, while the hypochlorite concentration remains constant, the iodide concentration is essentially tripled { = } and the initial rate is essentially tripled { = }. This indicates a first order with respect to the iodide ion.

(ii) comparing expt. 1 to expt. 3, while the iodide concentration remains essentially constant (a 2.7% drop), the hypochlorite concentration is essentially quadrupled { = } and the initial rate is essentially quadrupled { = }. This indicates a first order with respect to the hypochlorite ion.

(i) from experiments 1 & 2

3.05 = = 3.1m, where m = 1

(ii) from experiments 1 & 3

3.82 = (0.94)

4.06 = 4.1n, where n = 1

(b)(i) rate = k[I–] [ClO–]

(ii) k = = = 610 L mol –1 s –1

(c) (i) vertical axis is “ln of [H2O2]”

(ii) units for k are min–1

(iii)

2004 B

The first-order decomposition of a colored chemical species, X, into colorless products is monitered with a spectrophotometer by measuring changes in absorbance over time. Species X has a molar absorptivity constant of 5.00103 cm–1M–1 and the pathlength of the cuvetee containing the reaction mixture is 1.00 cm. The data from the experiment are given in the table below.

[X]
(M) / Absorbance / Time
(min)
? / 0.600 / 0.0
4.0010–5 / 0.200 / 35.0
3.0010–5 / 0.150 / 44.2
1.5010–5 / 0.075 / ?

(a)Calculate the initial concentration of the unknown species.

(b)Calculate the rate constant for the first order reaction using the values given for concentration and time. Include units with your answers.

(c)Calculate the minutes it takes for the absorbance to drop from 0.600 to 0.075.

(d)Calculate the half-life of the reaction. Include units with your answer.

(e)Experiments were performed to determine the value of the rate constant for this reaction at various temperatures. Data from these experiments were used to produce the graph below, where T is temperature. This graph can be used to determine Ea, the activation energy.

(i)Label the vertical axis of the graph

(ii)Explain how to calculate the activation energy from this graph.

Answer:

(a)A = abc; 0.600 = (5000 cm–1M–1)(1.00 cm)(c)

c = 1.2010–4 M

(b)ln[X]t – ln[X]0 = –kt

ln(4.0010–5) – ln(1.2010–4) = –k(35 min)

k = 0.0314 min–1

ln[1.5010–5] – ln[1.2010–4] = –0.0314 min–1t

t = 66.2 min.

(d)t1/2 = = = 22.1 min

(e) (i)

(ii)= slope of the line, multiply the slope by –R to obtain Ea

2003 B

5 Br–(aq) + BrO3–(aq) + 6 H+(aq) 3 Br2(l) + 3 H2O(l)

In a study of the kinetics of the reaction represented above, the following data were obtained at 298 K.

Experiment / Initial [Br–] (mol L-1) / Initial [BrO3–] (mol L-1) / Initial [H+] (mol L-1) / Rate of Disappearance of BrO3– (mol L-1 s-1)
1 / 0.00100 / 0.00500 / 0.100 / 2.5010-4
2 / 0.00200 / 0.00500 / 0.100 / 5.0010-4
3 / 0.00100 / 0.00750 / 0.100 / 3.7510-4
4 / 0.00100 / 0.01500 / 0.200 / 3.0010-3

(a)From the data given above, determine the order of the reaction for each reactant listed below. Show your reasoning.

(i)Br–

(ii)BrO3–

(iii)H+

(b)Write the rate law for the overall reaction.

(c)Determine the value of the specific rate constant for the reaction at 298 K. Include the correct units.

(d)Calculate the value of the standard cell potential, E˚, for the reaction using the information in the table below.

Half-reaction / E˚ (V)
Br2(l) + 2e-  2 Br–(aq) / +1.065
BrO3–(aq) + 6 H+(aq) + 5e- 1/2 Br2(l) + 3 H2O(l) / +1.52

(e)Determine the total number of electrons transferred in the overall reaction.

2003 B

A rigid 5.00 L cylinder contains 24.5 g of N2(g) and 28.0 g of O2(g)

(a)Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K.

(b)The temperature of the gas mixture in the cylinder is decreased to 280 K. Calculate each of the following.

(i)The mole fraction of N2(g) in the cylinder.

(ii)The partial pressure, in atm, of N2(g) in the cylinder.

(c)If the cylinder develops a pinhole-sized leak and some of the gaseous mixture escapes, would the ratio in the cylinder increase, decrease, or remain the same? Justify your answer.

A different rigid 5.00 L cylinder contains 0.176 mol of NO(g) at 298 K. A 0.176 mol sample of O2(g) is added to the cylinder, where a reaction occurs to produce NO2(g).

(d)Write the balanced equation for the reaction.

(e)Calculate the total pressure, in atm, in the cylinder at 298 K after the reaction is complete.

Answer:

(a)24.5 g N2 = 0.875 mol N2

28.0 g O2 = 0.875 mol O2

P = =

= 8.56 atm

(b)(i) = 0.500 mole fraction N2

(ii)

= 8.05 atm  mole fraction = 8.05 atm  0.500

= 4.02 atm N2

(c)decrease; since N2 molecules are lighter than O2 they have a higher velocity and will escape more frequently (Graham’s Law), decreasing the amount of N2 relative to O2

(d)2 NO + O2 2 NO2

(e)all 0.176 mol of NO will react to produce 0.176 mol of NO2, only 1/2 of that amount of O2 will react, leaving 0.088 mol of O2, therefore, 0.176 + 0.088 = 0.264 mol of gas is in the container.

P = =

= 1.29 atm

2002 B

Answer parts (a) through (e) below, which relate to reactions involving silver ion, Ag+.

The reaction between silver ion and solid zinc is represented by the following equation.

2 Ag+(aq) + Zn(s) Zn2+(aq) + 2 Ag(s)

(a)A 1.50 g sample of Zn is combined with 250. mL of 0.110 M AgNO3 at 25˚C.

(i)Identify the limiting reactant. Show calculations to support your answer.

(ii)On the basis of the limiting reactant that you identified in part (i), determine the value of [Zn2+] after the reaction is complete. Assume that volume change is negligible.

(b)Determine the value of the standard potential, E˚, for a galvanic cell based on the reaction between AgNO3(aq) and solid Zn at 25˚C.

Another galvanic cell is based on the reaction between Ag+(aq) and Cu(s), represented by the equation below. At 25˚C, the standard potential, E˚, for the cell is 0.46 V.

2 Ag+(aq) + Cu(s) Cu2+(aq) + 2 Ag(s)

(c)Determine the value of the standard free-energy change, ∆G˚, for the reaction between Ag+(aq) and Cu(s) at 25˚C.

(d)The cell is constructed so that [Cu2+] is 0.045 M and [Ag+] is 0.010 M. Calculate the value of the potential, E, for the cell.

(e)Under the conditions specified in part (d), is the reaction in the cell spontaneous? Justify your answer.

Answer:

(a)(i) AgNO3 solution

1.50 g Zn   = 417 mL of silver nitrate solution required to completely react the zinc, therefore, AgNO3 is the limiting reagent.

(ii) 250 mL AgNO3  = 0.01375 mol Zn

= 0.0550 M [Zn2+]

(b)2 Ag+ + 2e- 2 AgE˚ = +0.80 v

Zn – 2e- Zn2+E˚ = +0.76 v

+1.56 v

(c)∆G˚ = -nFE˚ = -(2)(96500)(0.46 v) = -89000 J

(d)Ecell = E˚ - log = 0.46 - log= 0.38 v

(e)yes; any reaction is spontaneous with a positive voltage

2001 B

2 NO(g) + O2(g) 2 NO2(g)H°= -114.1 kJ, S°= -146.5 J K-1

The reaction represented above is one that contributes significantly to the formation of photochemical smog.

(a)Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g).

(b)For the reaction at 25C, the value of the standard free-energy change, G, is -70.4 kJ.

(i)Calculate the value of the equilibrium constant, Keq, for the reaction at 25C.

(ii)Indicate whether the value of G would become more negative, less negative, or remain unchanged as the temperature is increased. Justify your answer.

(c)Use the data in the table below to calculate the value of the standard molar entropy, S, for O2(g) at 25C.

Standard Molar Entropy, S (J K-1 mol-1)
NO(g) / 210.8
NO2(g) / 240.1

(d)Use the data in the table below to calculate the bond energy, in kJ mol-1, of the nitrogen-oxygen bond in NO2 . Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same energy).

Bond Energy (kJ mol-1)
Nitrogen-oxygen bond in NO / 607
Oxygen-oxygen bond in O2 / 495
Nitrogen-oxygen bond in NO2 / ?

Answer:

(a)73.1 g  = 139 kJ

(b)(i) Keq = e–G/RT = e–(–70400/(8.31)(298)) = 2.221012

(ii) less negative; G = H – TS; as temperature increases, –TS becomes a larger positive value causing an increase in G (less negative).

(c)S = S(products) – S(reactants)

-146.5 = [(2)(240.1)] – [(210.8)(2)+ Soxygen] J/K

Soxygen = +205.1 J/K

(d)2 NO(g) + O2(g) 2 NO2(g) + 114.1 kJ

H = enthalpy of bonds broken – enthalpy of bonds formed

-114.1 = [(607)(2) + 495] - 2X

X = 912 kJ / 2 N=O bonds

456 kJ = bond energy for N=O bond

2001 B

Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin.

(a)The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet.

(b)The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The combustion of 3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry carbon dioxide, measured at 750. mm Hg and 25C. Calculate the mass, in g, of each element in the 3.000 g sample.

(c)A student dissolved 1.625 g of pure acetylsalicylic acid in distilled water and titrated the resulting solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming that acetylsalicylic acid has only one ionizable hydrogen, calculate the molar mass of the acid.

(d)A 2.00×10–3 mole sample of pure acetylsalicylic acid was dissolved in 15.00 mL of water and then titrated with 0.100 M NaOH(aq). The equivalence point was reached after 20.00 mL of the NaOH solution had been added. Using the data from the titration, shown in the table below, determine

(i)the value of the acid dissociation constant, Ka, for acetylsalicylic acid and

(ii)the pH of the solution after a total volume of 25.00 mL of the NaOH solution had been added (assume that volumes are additive).

Volume of 0.100M NaOH Added (mL) / pH
0.00 / 2.22
5.00 / 2.97
10.00 / 3.44
15.00 / 3.92
20.00 / 8.13
25.00 / ?

Answer:

(a) 100% = 16.3%

(b)1.200 g H2O  + 16 g H2O) = 0.134 g H

n = = = 0.150 mol CO2

0.150 mol CO2 = 1.801 g C

3.000 g ASA – (1.801 g C + 0.134 g H) = 1.065 g O

1 mol base = 1 mol acid

= 180 g/mol

(d)(i) HAsa  Asa– + H+

= 0.133 M

pH = –log[H+]; 2.22 = –log[H+]

[H+] = M = [Asa–]

[HAsa] = 0.133 M – 6.03  10-3 M = 0.127 M

K = = = 2.85 10-4

when the solution is half-neutralized, pH = pKa

at 10.00 mL, pH = 3.44; K = 10–pH

= 10–3.44 = 3.6310-4

(ii) 0.025 L  0.100 mol/L = 2.50  10-3 mol OH-

2.50  10-3 mol OH- - 2.00  10-3 mol neutralized = 5.0  10-4 mol OH- remaining in (25 + 15 mL) of solution; [OH-] = 5.010-4 mol/0.040 L = 0.0125 M

pH = 14 – pOH = 14 + log[OH-] = 14 – 1.9 = 12.1