Consider the titration of a 24.5 mL sample of 0.115 M RbOH with
0.110 M HCL . Determine each of the following:
A) initial pH
B) the volume of added acid required to reach the equivalence
point
C) the pH at 5.1 mL of added acid
D) the pH at the equivalence point
E) the pH after adding 5.6 mL of acid beyond the equivalence
point
A)when 0.0 mL of HCl is added
Given:
M(HCl) = 0.11 M
V(HCl) = 0 mL
M(RbOH) = 0.115 M
V(RbOH) = 24.5 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.11 M * 0 mL = 0 mmol
mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.115 M * 24.5 mL = 2.8175 mmol
We have:
mol(HCl) = 0 mmol
mol(RbOH) = 2.818 mmol
0 mmol of both will react
remaining mol of RbOH = 2.818 mmol
Total volume = 24.5 mL
[OH-]= mol of base remaining / volume
[OH-] = 2.818 mmol/24.5 mL
= 0.115 M
use:
pOH = -log [OH-]
= -log (0.115)
= 0.9393
use:
PH = 14 - pOH
= 14 - 0.9393
= 13.0607
Answer: 13.06
B)
M(RbOH)=0.115 M
M(HCl)=0.11 M
V(RbOH)=24.5 mL
According to balanced reaction:
1*number of mol of RbOH =1*number of mol of HCl
1*M(RbOH)*V(RbOH) =1*M(HCl)*V(HCl)
1*0.115 M *24.5 mL = 1*0.11M *V(HCl)
V(HCl) = 25.6 mL
Answer: 25.6 mL
C)when 5.1 mL of HCl is added
Given:
M(HCl) = 0.11 M
V(HCl) = 5.1 mL
M(RbOH) = 0.115 M
V(RbOH) = 24.5 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.11 M * 5.1 mL = 0.561 mmol
mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.115 M * 24.5 mL = 2.8175 mmol
We have:
mol(HCl) = 0.561 mmol
mol(RbOH) = 2.818 mmol
0.561 mmol of both will react
remaining mol of RbOH = 2.257 mmol
Total volume = 29.6 mL
[OH-]= mol of base remaining / volume
[OH-] = 2.257 mmol/29.6 mL
= 7.623*10^-2 M
use:
pOH = -log [OH-]
= -log (7.623*10^-2)
= 1.1179
use:
PH = 14 - pOH
= 14 - 1.1179
= 12.8821
Answer: 12.88
D)
This is titration of strong acid and strong base.
pH at equivalence point will be 7.00 as the solution would be neutral
Answer: 7.00
E)
1)when 31.2 mL of HCl is added
Given:
M(HCl) = 0.11 M
V(HCl) = 31.2 mL
M(RbOH) = 0.115 M
V(RbOH) = 24.5 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.11 M * 31.2 mL = 3.432 mmol
mol(RbOH) = M(RbOH) * V(RbOH)
mol(RbOH) = 0.115 M * 24.5 mL = 2.8175 mmol
We have:
mol(HCl) = 3.432 mmol
mol(RbOH) = 2.818 mmol
2.818 mmol of both will react
remaining mol of HCl = 0.6145 mmol
Total volume = 55.7 mL
[H+]= mol of acid remaining / volume
[H+] = 0.6145 mmol/55.7 mL
= 1.103*10^-2 M
use:
pH = -log [H+]
= -log (1.103*10^-2)
= 1.9573
Answer: 1.96
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