Question

Consider the titration of a 23.0 mL sample of 0.120 molL−1 RbOH with 0.110 molL−1 HCl. Determine each quantity:

Part A

the volume of added acid required to reach the equivalence point

Part B

the pH at 5.7 mL of added acid

Express your answer using two decimal places.

Part C

the pH after adding 6.0 mL of acid beyond the equivalence point

Express your answer using two decimal places.

Answer 1

A)

Balanced chemical equation is:

RbOH + HCl ---> RbCl + H2O

Here:

M(RbOH)=0.12 M

M(HCl)=0.11 M

V(RbOH)=23.0 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*0.12 M *23.0 mL = 1*0.11M *V(HCl)

V(HCl) = 25.1 mL

Answer: 25.1 mL

B)

1)when 5.7 mL of HCl is added

Given:

M(HCl) = 0.11 M

V(HCl) = 5.7 mL

M(RbOH) = 0.12 M

V(RbOH) = 23 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.11 M * 5.7 mL = 0.627 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.12 M * 23 mL = 2.76 mmol

We have:

mol(HCl) = 0.627 mmol

mol(RbOH) = 2.76 mmol

0.627 mmol of both will react

remaining mol of RbOH = 2.133 mmol

Total volume = 28.7 mL

[OH-]= mol of base remaining / volume

[OH-] = 2.133 mmol/28.7 mL

= 7.432*10^-2 M

use:

pOH = -log [OH-]

= -log (7.432*10^-2)

= 1.1289

use:

PH = 14 - pOH

= 14 - 1.1289

= 12.8711

Answer: 12.87

C)

Given:

M(HCl) = 0.11 M

V(HCl) = 31.1 mL

M(RbOH) = 0.12 M

V(RbOH) = 23 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.11 M * 31.1 mL = 3.421 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.12 M * 23 mL = 2.76 mmol

We have:

mol(HCl) = 3.421 mmol

mol(RbOH) = 2.76 mmol

2.76 mmol of both will react

remaining mol of HCl = 0.661 mmol

Total volume = 54.1 mL

[H+]= mol of acid remaining / volume

[H+] = 0.661 mmol/54.1 mL

= 1.222*10^-2 M

use:

pH = -log [H+]

= -log (1.222*10^-2)

= 1.913

Answer: 1.913