Question

estion 3 of 9 Calculate the pH for each of the cases in the titration of 35.0 mL of 0.220 M KOH(aq) with 0.220M HI(aq). Note: Enter your answers with two decimal places. after addition of 13.5 mL HI: before addition of any HI: after addition of 20.5 mL HI: after addition of 35.0 mL HI: after addition of 42.5 mL HI: after addition of 50.0 mL HI:

Answer 1

1)when 0.0 mL of HI is added

Given:

M(HI) = 0.22 M

V(HI) = 0 mL

M(KOH) = 0.22 M

V(KOH) = 35 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.22 M * 0 mL = 0 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.22 M * 35 mL = 7.7 mmol

We have:

mol(HI) = 0 mmol

mol(KOH) = 7.7 mmol

0 mmol of both will react

remaining mol of KOH = 7.7 mmol

Total volume = 35.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 7.7 mmol/35.0 mL

= 0.22 M

use:

pOH = -log [OH-]

= -log (0.22)

= 0.6576

use:

PH = 14 - pOH

= 14 - 0.6576

= 13.3424

Answer: 13.34

2)when 13.5 mL of HI is added

Given:

M(HI) = 0.22 M

V(HI) = 13.5 mL

M(KOH) = 0.22 M

V(KOH) = 35 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.22 M * 13.5 mL = 2.97 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.22 M * 35 mL = 7.7 mmol

We have:

mol(HI) = 2.97 mmol

mol(KOH) = 7.7 mmol

2.97 mmol of both will react

remaining mol of KOH = 4.73 mmol

Total volume = 48.5 mL

[OH-]= mol of base remaining / volume

[OH-] = 4.73 mmol/48.5 mL

= 9.753*10^-2 M

use:

pOH = -log [OH-]

= -log (9.753*10^-2)

= 1.0109

use:

PH = 14 - pOH

= 14 - 1.0109

= 12.9891

Answer: 12.99

3)when 20.5 mL of HI is added

Given:

M(HI) = 0.22 M

V(HI) = 20.5 mL

M(KOH) = 0.22 M

V(KOH) = 35 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.22 M * 20.5 mL = 4.51 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.22 M * 35 mL = 7.7 mmol

We have:

mol(HI) = 4.51 mmol

mol(KOH) = 7.7 mmol

4.51 mmol of both will react

remaining mol of KOH = 3.19 mmol

Total volume = 55.5 mL

[OH-]= mol of base remaining / volume

[OH-] = 3.19 mmol/55.5 mL

= 5.748*10^-2 M

use:

pOH = -log [OH-]

= -log (5.748*10^-2)

= 1.2405

use:

PH = 14 - pOH

= 14 - 1.2405

= 12.7595

Answer: 12.76

4)when 35.0 mL of HI is added

Given:

M(HI) = 0.22 M

V(HI) = 35 mL

M(KOH) = 0.22 M

V(KOH) = 35 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.22 M * 35 mL = 7.7 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.22 M * 35 mL = 7.7 mmol

We have:

mol(HI) = 7.7 mmol

mol(KOH) = 7.7 mmol

7.7 mmol of both will react to form neutral solution

hence pH of solution will be 7

Answer: 7.00

5)when 42.5 mL of HI is added

Given:

M(HI) = 0.22 M

V(HI) = 42.5 mL

M(KOH) = 0.22 M

V(KOH) = 35 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.22 M * 42.5 mL = 9.35 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.22 M * 35 mL = 7.7 mmol

We have:

mol(HI) = 9.35 mmol

mol(KOH) = 7.7 mmol

7.7 mmol of both will react

remaining mol of HI = 1.65 mmol

Total volume = 77.5 mL

[H+]= mol of acid remaining / volume

[H+] = 1.65 mmol/77.5 mL

= 2.129*10^-2 M

use:

pH = -log [H+]

= -log (2.129*10^-2)

= 1.6718

Answer: 1.67

6)when 50.0 mL of HI is added

Given:

M(HI) = 0.22 M

V(HI) = 50 mL

M(KOH) = 0.22 M

V(KOH) = 35 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.22 M * 50 mL = 11 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.22 M * 35 mL = 7.7 mmol

We have:

mol(HI) = 11 mmol

mol(KOH) = 7.7 mmol

7.7 mmol of both will react

remaining mol of HI = 3.3 mmol

Total volume = 85.0 mL

[H+]= mol of acid remaining / volume

[H+] = 3.3 mmol/85.0 mL

= 3.882*10^-2 M

use:

pH = -log [H+]

= -log (3.882*10^-2)

= 1.4109

Answer: 1.41