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Use the References to access important values if needed for this question. Determine the pH during the titration of 23.5 mL oUse the References to access important values if needed for this question. Determine the pH during the titration of 28.9 mL o

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Answer #1

1)when 0.0 mL of KOH is added

Given:

M(HNO3) = 0.17 M

V(HNO3) = 23.5 mL

M(KOH) = 0.109 M

V(KOH) = 0 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.17 M * 23.5 mL = 3.995 mmol

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

mol(KOH) = 0.109 M * 0 mL = 0 mmol

We have:

mol(HNO3) = 3.995 mmol

mol(KOH) = 0 mmol

0 mmol of both will react

remaining mol of HNO3 = 3.995 mmol

Total volume = 23.5 mL

[H+]= mol of acid remaining / volume

[H+] = 3.995 mmol/23.5 mL

= 0.17 M

use:

pH = -log [H+]

= -log (0.17)

= 0.7696

Answer: 0.7696

2)when 18.3 mL of KOH is added

Given:

M(HNO3) = 0.17 M

V(HNO3) = 23.5 mL

M(KOH) = 0.109 M

V(KOH) = 18.3 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.17 M * 23.5 mL = 3.995 mmol

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

mol(KOH) = 0.109 M * 18.3 mL = 1.9947 mmol

We have:

mol(HNO3) = 3.995 mmol

mol(KOH) = 1.995 mmol

1.995 mmol of both will react

remaining mol of HNO3 = 2 mmol

Total volume = 41.8 mL

[H+]= mol of acid remaining / volume

[H+] = 2 mmol/41.8 mL

= 4.785*10^-2 M

use:

pH = -log [H+]

= -log (4.785*10^-2)

= 1.3201

Answer: 1.32

3)

This is titration of strong acid and strong base.

At equivalence point, solution would be neutral as equal mol of acid and base would have been added.

So, pH would be 7.00

Answer: 7.00

4)when 44.9 mL of KOH is added

Given:

M(HNO3) = 0.17 M

V(HNO3) = 23.5 mL

M(KOH) = 0.109 M

V(KOH) = 44.9 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.17 M * 23.5 mL = 3.995 mmol

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

mol(KOH) = 0.109 M * 44.9 mL = 4.8941 mmol

We have:

mol(HNO3) = 3.995 mmol

mol(KOH) = 4.894 mmol

3.995 mmol of both will react

remaining mol of KOH = 0.8991 mmol

Total volume = 68.4 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.8991 mmol/68.4 mL

= 1.314*10^-2 M

use:

pOH = -log [OH-]

= -log (1.314*10^-2)

= 1.8812

use:

PH = 14 - pOH

= 14 - 1.8812

= 12.1188

Answer: 12.12

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