1) What is the pH at the equivalence point in the titration of a 24.7 mL sample
ID: 1032619 • Letter: 1
Question
1) What is the pH at the equivalence point in the titration of a 24.7 mL sample of a 0.332 M aqueous acetic acid solution with a 0.470 M aqueous potassium hydroxide solution? pH =
2) A 42.5 mL sample of a 0.568 M aqueous hypochlorous acid solution is titrated with a 0.399 M aqueous barium hydroxide solution. What is the pH after 21.0 mL of base have been added? pH =
3) When a 18.9 mL sample of a 0.365 M aqueous hydrocyanic acid solution is titrated with a 0.328 M aqueous sodium hydroxide solution, what is the pH after 31.5 mL of sodium hydroxide have been added? pH
Explanation / Answer
1)
acetic acid= 24.7 ml of 0.332 M
number of moles of acetic acid= 0.332Mx0.0247L= 0.0082 moles
KOH= 0.470M
at equivalent point volume of KOH = 0.332x24.7/0.470 = 17.45 ml
Total volume of the solution = 24.7 + 17.45 = 42.15 ml = 0.04215L
at equivalent point
PH= 7 + 1/2[PKa + logC]
Pka of acetic aci= 4.75
C= number of moles /total volume= 0.0082/0.04215 = 0.1945 M
PH= 7 + 1/2[4.75 + log(0.1945)]
PH= 9.02
2)
HCl = 42.5 ml of 0.568M
number of moles of HCl= 0.568M x0.0425L= 0.02414 moles
Ba(OH)2 = 0.399M of 21.0 ml
number of moles of Ba(OH)2 = 0.399Mx0.021L=0.008379 moles
number o fmoles of OH- = 2x0.008379 =0.01676 moles
number of moles of HCl is greater than the Ba(OH)2
remaining number of moles of acid = 0.02414 - 0.01676= 0.00738 moles
number of moles of H+ = 0.00738 moles
total volume = 42.5 + 21.0 = 63.5 mL = 0.0635L
[H+]= number of moles /total volume = 0.00738/0.0635 =0.1162
[H+} = 0.1162M
-log(H+) = -log(0.1162)
PH= 0.935M.
3)
HCN= 18.9 ml of 0.365M
number of moles of HCN= 0.365Mx0.0189L=0.006899 moles
NaOH= 31.5 ml of 0.328M
number of moles of NaOH= 0.328Mx0.0315L=0.01033 moles
number o fmoles of Base is greater thanthe HCN. so the buffer solution is not possible
remaining numberr of moles of NaOH= 0.01033 - 0.006899 = 0.003431moles
number of moles of OH- = 0.003431 moles
total volume = 18.9 + 31.5 = 50.4 ml = 0.0504L
[OH-] = 0.003431/0.0504 = 0.0681 M
-log[OH-] = -log0.0681)
POH= 1.167
PH + POH= 14
PH= 14 - POH
PH= 14- 1.167
PH= 12.83.
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