Quiz #3. Chem 1007, Summer 2014 Name onsider the weak acid with a Ka = 5.0 x 10-
ID: 1018460 • Letter: Q
Question
Quiz #3. Chem 1007, Summer 2014 Name onsider the weak acid with a Ka = 5.0 x 10-5. Consiger the weak aoid, ascobic adid, HAs Htf Oa)betterknown as vitamin C. a. Calculate the initial pH of a 0.100 M HAsc solution. (5 pts) 1 o.loo o-oce C.oCo A) (A -1% 2.22/i>, 2.66 b. What would be the pH if 10 mL of a 0.125 M solution of sodium hydroxide were titrated to 50-mL of the original 0.100 M HAsc solution. (3 pts) -0625 -125 1-253 c. What would be the pH if 15 ml of a 0.125 M solution of sodium hydroxide were titrated to 50-mL of the original 0.100 M HAsc solution. (3 pts) 0-124M d. What is the pH if 20 mL of a 0.125 M solution of sodium hydroxide were titrated to 50-mL of the original 0.100 M HAsc solution. (3 pts)Explanation / Answer
Solution :-
HAsc Ka= 5.0*10^-5
pka = - log ka
pka = -log 5.0*10^-5
= 4.30
Part a) Initial pH of the 0.100 M HAsc
HASc + H2O ------ > H3O^+ + Asc^-
0.100 0 0
-x +x +x
0.100-x x x
Ka = [H3O+][Asc-]/[HAsc]
5.0*10^-5 = [x][x]/[0.100-x]
Since the ka is very small we can neglect the x from denominator
5.0*10^-5 = [x][x]/[0.100]
5.0*10^-5 * 0.100 = x^2
5.0*10^-6 = x^2
Taking square root of both sides we get
2.236*10^-3 = x
pH= -log [H3O+]
pH= -log [2.236*10^-3]
pH= 2.65
part b ) pH after adding 10 ml of 0.125 M NaOH to 50 ml of 0.100 M HAsc
initial moles of HAsc = 0.100 mol per L * 0.050 L= 0.005 mol
initial moles of NaOH = 0.125 mol per L * 0.010 L = 0.00125 mol NaOH
After the reaction moles of HAsc remain = 0.005 mol – 0.00125 mol = 0.00375 mol
Moles of conjugate base Asc- formed = 0.00125 mol
Now calculating the pH
pH= pka + log [conju .Base]/[acid]
pH= 4.30 + log [0.00125 / 0.00375]
pH= 3.82
part c) after adding 15 ml of 0.125 M NaOH
initial moles of HAsc = 0.100 mol per L * 0.050 L= 0.005 mol
initial moles of NaOH = 0.125 mol per L * 0.015 L = 0.001875 mol NaOH
After the reaction moles of HAsc remain = 0.005 mol – 0.001875 mol = 0.003125 mol
Moles of conjugate base Asc- formed = 0.001875 mol
Now calculating the pH
pH= pka + log [conju .Base]/[acid]
pH= 4.30 + log [0.001875 / 0.003125]
pH= 4.08
part d) After adding 20 ml of 0.125 M NaOH
initial moles of HAsc = 0.100 mol per L * 0.050 L= 0.005 mol
initial moles of NaOH = 0.125 mol per L * 0.020 L = 0.0025 mol NaOH
After the reaction moles of HAsc remain = 0.005 mol – 0.0025 mol = 0.0025 mol
Moles of conjugate base Asc- formed = 0.0025 mol
Now calculating the pH
pH= pka + log [conju .Base]/[acid]
pH= 4.30 + log [0.0025 / 0.0025]
pH= 4.30
part e ) After adding 38 ml of 0.125 M NaOH
initial moles of HAsc = 0.100 mol per L * 0.050 L= 0.005 mol
initial moles of NaOH = 0.125 mol per L * 0.038 L = 0.00475 mol NaOH
After the reaction moles of HAsc remain = 0.005 mol – 0.00475 mol = 0.00025 mol
Moles of conjugate base Asc- formed = 0.00475 mol
Now calculating the pH
pH= pka + log [conju .Base]/[acid]
pH= 4.30 + log [0.00475 / 0.00025]
pH= 5.58
part f) After adding 40 ml of 0.125 M NaOH
initial moles of HAsc = 0.100 mol per L * 0.050 L= 0.005 mol
initial moles of NaOH = 0.125 mol per L * 0.040 L = 0.005 mol NaOH
After the reaction moles of HAsc remain = 0.005 mol – 0.005 mol = 0.00 mol
Moles of conjugate base Asc- formed = 0.005 mol
New molarity of conjugate base = 0.005 mol / 0.090 L =0.055 M
(total volume = 50 + 40 = 90 ml = 0.090 L)
Conjugate base reacts with water
Asc^- + H2O -------- > HAsc + OH-
0.055 0 0
-x +x +x
0.055-x x x
Kb= [HAsc][OH-]/[Asc-]
(1*10^-14/5.0*10^-5) = [x][x]/[0.055-x]
2*10^-10 = x^2 / 0.055
2*10^-10 * 0.055 = x^2
1.1*10^-11 = x^2
Taking square root of both sides we get
3.32*10^-6 = x
pOH = -log [OH-]
pOH = - log [3.32*10^-6]
pOH= 5.47
pH= 14 – pOH
pH= 14 – 5.47
pH= 8.53
part g )
pH after adding 50 ml of 0.125 M NaOH
Moles of NaOH = 0.125 mol per L * 0.050 L = 0.00625 mol
Moles of HAsc = 0.100 mol per L * 0.050 L = 0.005 mol
So after the reaction moles of NaOH remain = 0.00625 mol – 0.0050 mol = 0.00125 mol
Total volume = 50 ml + 50 ml = 100 ml
New molarity of the OH- = 0.00125 mol / 0.100 L = 0.0125 M
pOH= -log [OH-]
pOH= - log [0.0125 ]
pOH= 1.90
pH= 14 – pOH
pH= 14 – 1.90
pH= 12.1