Question

consider the titration of a 34.0 mL sample of a 0.180 M HBr with 0.210 M KOH. determine the following:
a. initial pH
b. the volume if added base required to reach the equivalence point
c. the pH at 10.6 mL of added base
d. the pH at the equivalence point
e. the pH after adding 5.0 mL of base beyond the equivalence point

Answer 1

a)when 0.0 mL of KOH is added
Given:
M(HBr) = 0.18 M
V(HBr) = 34 mL
M(KOH) = 0.21 M
V(KOH) = 0 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.18 M * 34 mL = 6.12 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.21 M * 0 mL = 0 mmol


We have:
mol(HBr) = 6.12 mmol
mol(KOH) = 0 mmol
0 mmol of both will react
remaining mol of HBr = 6.12 mmol
Total volume = 34.0 mL

[H+]= mol of acid remaining / volume
[H+] = 6.12 mmol/34 mL
= 0.18 M


use:
pH = -log [H+]
= -log (0.18)
= 0.7447
Answer: 0.745

b)
Balanced chemical equation is:
HBr + KOH ---> KBr + H2O


Here:
M(HBr)=0.18 M
M(KOH)=0.21 M
V(HBr)=34.0 mL

According to balanced reaction:
1*number of mol of HBr =1*number of mol of KOH
1*M(HBr)*V(HBr) =1*M(KOH)*V(KOH)
1*0.18 M *34.0 mL = 1*0.21M *V(KOH)
V(KOH) = 29.1429 mL

Answer: 29.1 mL

C)when 10.6 mL of KOH is added
Given:
M(HBr) = 0.18 M
V(HBr) = 34 mL
M(KOH) = 0.21 M
V(KOH) = 10.6 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.18 M * 34 mL = 6.12 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.21 M * 10.6 mL = 2.226 mmol


We have:
mol(HBr) = 6.12 mmol
mol(KOH) = 2.226 mmol
2.226 mmol of both will react
remaining mol of HBr = 3.894 mmol
Total volume = 44.6 mL

[H+]= mol of acid remaining / volume
[H+] = 3.894 mmol/44.6 mL
= 8.731*10^-2 M


use:
pH = -log [H+]
= -log (8.731*10^-2)
= 1.0589
Answer: 1.06

d)
This is titration of strong acid and strong base.
So, the solution would be neutral at equivalence point.
Hence pH would be 7.00
Answer: 7.00

e)

1)when 34.1 mL of KOH is added
Given:
M(HBr) = 0.18 M
V(HBr) = 34 mL
M(KOH) = 0.21 M
V(KOH) = 29.1 mL + 5.00 mL = 34.1 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.18 M * 34 mL = 6.12 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.21 M * 34.1 mL = 7.161 mmol


We have:
mol(HBr) = 6.12 mmol
mol(KOH) = 7.161 mmol
6.12 mmol of both will react

remaining mol of KOH = 1.041 mmol
Total volume = 68.1 mL

[OH-]= mol of base remaining / volume
[OH-] = 1.041 mmol/68.1 mL
= 1.529*10^-2 M


use:
pOH = -log [OH-]
= -log (1.529*10^-2)
= 1.8157


use:
PH = 14 - pOH
= 14 - 1.8157
= 12.1843
Answer: 12.18