Question

Calculate the pH when the following quantities of 0.100 M HNO₃ have been added to 25.00 mL of 0.100 M KOH solution:

1. Initial pH

2. pH at 24.90 mL of HNO₃ added

3. pH at Equivalence point (25.00 mL of HNO₃ added)

4.pH at 25.10 mL of HNO₃ added

5. Final pH (good rule of thumb- 2x volume added to reach equivalence point 50.00 mL of HNO₃ added)

6. Sketch the titration curve for this reaction using the above 5 points

Answer 1

1)

Initially there is only KOH

[OH-] = 0.100 M

use:

pOH = -log [OH-]

= -log (0.1)

= 1

use:

PH = 14 - pOH

= 14 - 1

= 13

Answer: 13.0

2)

Given:

M(HNO3) = 0.1 M

V(HNO3) = 24.9 mL

M(KOH) = 0.1 M

V(KOH) = 25 mL

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

mol(HNO3) = 0.1 M * 24.9 mL = 2.49 mmol

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

mol(KOH) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HNO3) = 2.49 mmol

mol(KOH) = 2.5 mmol

2.49 mmol of both will react

remaining mol of KOH = 0.01 mmol

Total volume = 49.9 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.01 mmol/49.9 mL

= 0.0002 M

use:

pOH = -log [OH-]

= -log (2.004*10^-4)

= 3.6981

use:

PH = 14 - pOH

= 14 - 3.6981

= 10.3019

Answer: 10.30

3)

At equivalence point, solution will be neutral as this is titration of strong acid and strong base

So, pH will be 7

Answer: 7.00

4)

Given:

M(HNO3) = 0.1 M

V(HNO3) = 25.1 mL

M(KOH) = 0.1 M

V(KOH) = 25 mL

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

mol(HNO3) = 0.1 M * 25.1 mL = 2.51 mmol

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

mol(KOH) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HNO3) = 2.51 mmol

mol(KOH) = 2.5 mmol

2.5 mmol of both will react

remaining mol of HNO3 = 0.01 mmol

Total volume = 50.1 mL

[H+]= mol of acid remaining / volume

[H+] = 0.01 mmol/50.1 mL

= 0.0002 M

use:

pH = -log [H+]

= -log (1.996*10^-4)

= 3.6998

Answer: 3.70

only 4 parts at a time