Consider that HI is a strong acid and that KOH is a strong base. For each case,

Question

Consider that HI is a strong acid and that KOH is a strong base. For each case, determine whether HI or KOH is in excess before calculating the pH.

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.200 M KOH(aq), with 0.200 M HI(aq) Number Note: Enter your answers with two decimal places. (a) before addition of any Hl Number (b) after addition of 13.5 mL of Hl Number (c) after addition of 22.5 mL of HI Number (d) after the addition of 35.0 mL of HII Number (e) after the addition of 43.5 mL of HI Number (f) after the addition of 50.0 mL of HI

Explanation / Answer

Let us write the balanced equation for the reaction

HI + KOH --> KI +H2O

A) At 0ml of HI

[OH-] = 0.2 M

pOH = -log[OH-] =0.6989
and
pH = 14-pOH = 13.30

b) At 13.5 ml of HI

No. of moles of HI = 0.2 M x 0.0135 L = 0.0027 moles
No. of moles of KOH = 0.2M x 0.035 L = 0.007 moles

No. of moles of KOH is larger than moles of HI,
[OH-] = [(moles KOH)-(moles HI)]/(total liters) = (0.007-0.0027)/0.0485 = 0.08866 M
pOH = -log[OH-] =1.052
and
pH = 14-pOH = 12.95

c) At 22.5 ml of HI

No. of moles of HI = 0.2 M x 0.0225 L = 0.0045 moles
No. of moles of KOH = 0.2M x 0.035 L = 0.007 moles

No. of moles of KOH is larger than moles of HI,
[OH-] = [(moles KOH)-(moles HI)]/(total liters) = (0.0025)/0.0485 = 0.04347 M
pOH = -log[OH-] =1.3617
and
pH = 14-pOH = 12.64

d) At 35 ml of HI

No. of moles of HI = 0.2 M x 0.035 L = 0.007 moles
No. of moles of KOH = 0.2M x 0.035 L = 0.007 moles

No. of moles of KOH is equal to moles of HI,
Hence pH = 7.00

e) At 43.5 ml of HI

No. of moles of HI = 0.2 M x 0.0435 L = 0.0087 moles
No. of moles of KOH = 0.2M x 0.035 L = 0.007 moles

No. of moles of KOH is smallr than moles of HI,
[H+] = [(moles HI)-(moles KOH)]/(total liters) = (0.0017)/0.0785 = 0.02165 M
pH = -log[H+] = 1.66

f) At 50 ml of HI

No. of moles of HI = 0.2 M x 0.050 L = 0.01 moles
No. of moles of KOH = 0.2M x 0.035 L = 0.007 moles

No. of moles of KOH is smallr than moles of HI,
[H+] = [(moles HI)-(moles KOH)]/(total liters) = (0.003)/0.085 = 0.03529 M
pH = -log[H+] = 1.45

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