Question

Part A:

Consider the titration of a 28.7 mL sample of 0.117 M RbOH with 0.186 M HCl.

Calculate the volume of added acid required to reach the equivalence point.
Express your answer to 3 significant figures.

Part B:

Calculate the pH after 8.2 mL of acid have been added.
Express your answer to 2 decimal places.

Part C:

Calculate the pH after 8.2 mL of acid have been added beyond the equivalence point.
Express your answer to 2 decimal places.

Answer 1

A)

Balanced chemical equation is:

RbOH + HCl ---> RbCl + H2O

Here:

M(RbOH)=0.117 M

M(HCl)=0.186 M

V(RbOH)=28.7 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*0.117 M *28.7 mL = 1*0.186M *V(HCl)

V(HCl) = 18.1 mL

Answer: 18.1 mL

B)

Given:

M(HCl) = 0.186 M

V(HCl) = 8.2 mL

M(RbOH) = 0.117 M

V(RbOH) = 28.7 mL

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

mol(HCl) = 0.186 M * 8.2 mL = 1.5252 mmol

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

mol(RbOH) = 0.117 M * 28.7 mL = 3.3579 mmol

We have:

mol(HCl) = 1.525 mmol

mol(RbOH) = 3.358 mmol

1.525 mmol of both will react

remaining mol of RbOH = 1.833 mmol

Total volume = 36.9 mL

[OH-]= mol of base remaining / volume

[OH-] = 1.833 mmol/36.9 mL

= 4.967*10^-2 M

use:

pOH = -log [OH-]

= -log (4.967*10^-2)

= 1.3039

use:

PH = 14 - pOH

= 14 - 1.3039

= 12.6961

Answer: 12.70

C)

Given:

M(HCl) = 0.186 M

V(HCl) = 26.3 mL

M(RbOH) = 0.117 M

V(RbOH) = 28.7 mL

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

mol(HCl) = 0.186 M * 26.3 mL = 4.8918 mmol

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

mol(RbOH) = 0.117 M * 28.7 mL = 3.3579 mmol

We have:

mol(HCl) = 4.892 mmol

mol(RbOH) = 3.358 mmol

3.358 mmol of both will react

remaining mol of HCl = 1.534 mmol

Total volume = 55.0 mL

[H+]= mol of acid remaining / volume

[H+] = 1.534 mmol/55.0 mL

= 2.789*10^-2 M

use:

pH = -log [H+]

= -log (2.789*10^-2)

= 1.5546

Answer: 1.55