Question

In: Chemistry

25.0 mL of a 0.100 M HCN solution is titrated with 0.100 M KOH. Ka of...

25.0 mL of a 0.100 M HCN solution is titrated with 0.100 M KOH. Ka of HCN=6.2 x 10^-10.
a)What is the pH when 25mL of KOH is added (this is the equivalent point).
b) what is the pH when 35 mL of KOH is added.

Solutions

Expert Solution

a) When 25.0mL of KOH is added

Molarity of HCN = 0.100M

Volume of HCN = 25.0 mL

No. of mol of HCN is

Molarity of KOH = 0.100M

When 25.0 mL of KOH is added, the no. of mol of HCN and that of KOH is the same.

So, the no. of mol of CN- formed will also be equal to the no. of mol of HCN and KOH.

Therefore, no. of mol of CN- = 2.50 x 10-3 mol

Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 50.0 x 10-3L

Concentration of CN- is

The CN- will undergo hydrolysis

The ICE table is

CN- H2O HCN OH-
I 0.0500M - 0 0
C -x - +x +x
E 0.0500-x - x x

The Kb of CN- 1.61 x 10-5 is very less, so the value of x will be very less such that it can be neglected.

Check if the approximation is valid or not.

If the % of [OH-] with respect to initial [CN-] is less than 5%, then the approximation of x is valid.

% whic is less than 5%

So, the approximation holds good.

Now, find pOH from [OH-]

Therefore, pH at equivalence point is 10.95

b) When 35.0mL of KOH is added.

Molarity of KOH = 0.100M

No. of mol of OH- taken is

No. of mol of HCN = 2.50 x 10-3mol

Excess no. of mol of OH- is

Total volume = 25.0 mL + 35.0 mL = 60.0 mL = 60.0 x 10-3L

Therefore, concentration of excess OH- is

Therefore, pH of the solution when 35.0 mL of KOH is added is 12.22


Related Solutions

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution....
A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL. (25 points)
A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution....
A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL. (25 points)
A 25.0−mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate...
A 25.0−mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution (a) 10.0 mL (b) 12.5 mL (c) 15.0 mL
A 25.0-mL solution of 0.100 M acetic acid is titrated with a 0.200 M KOH solution....
A 25.0-mL solution of 0.100 M acetic acid is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: a) 0.0 mL b) 5.0 mL c) 10.0 mL d) 12.5 mL e) 15.0 mL
A 20.00 mL solution of 0.100 M HCOOH (formic) was titrated with 0.100 M KOH. The...
A 20.00 mL solution of 0.100 M HCOOH (formic) was titrated with 0.100 M KOH. The Ka for the weak acid formic is 1.40 x 10-5. a. Determine the pH for the formic prior to its titration with KOH. b. Determine the pH of this solution at the ½ neutralization point of the titration. c. Identify the conjugate acid-base pair species at the ½ neutralization point.
A 25.0 mL sample of 0.100 M HC2H3O2 (Ka = 1.8 x 10^-5) is titrated with...
A 25.0 mL sample of 0.100 M HC2H3O2 (Ka = 1.8 x 10^-5) is titrated with a 0.100 M NaOH solution. What is the pH after the addition of 12.5 mL of NaOH?
A 83.0 mL sample of 0.0500 M HNO3 is titrated with 0.100 M KOH solution. Calculate...
A 83.0 mL sample of 0.0500 M HNO3 is titrated with 0.100 M KOH solution. Calculate the pH after the following volumes of base have been added. (a) 11.2 mL pH = _________ (b) 39.8 mL pH = __________    (c) 41.5 mL pH = ___________ (d) 41.9 mL pH = ____________    (e) 79.3 mL pH = _____________
25.0 mL of a 0.100 M NH3 is titrated with a strong acid. 0.100 M HCl....
25.0 mL of a 0.100 M NH3 is titrated with a strong acid. 0.100 M HCl. Calculate the pH of the NH3 solution at the following points during the titration: (Kb= 1.8 x 10^-5) A. Prior to the addition of any HCl. B: After the addition of 10.5 mL of a 0.100 M HCl. C: At the equivilance point. D: After the addition of 3 mL of 0.100 M HCl. Show your work.
Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C....
Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C. Ka for HCN = 6.2×10-10. Part 1 Calculate the pH after 0.0 mL of KOH has been added. pH = Part 2 Calculate the pH after 50.0 mL of KOH has been added. pH = Part 3 Calculate the pH after 75.0 mL of KOH has been added. pH = Part 4 Calculate the pH at the equivalence point. pH = Part 5 Calculate...
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M NaOH solution....
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M NaOH solution. Calculate the pH at each of the following points. Volume of NaOH added: 0, 5, 10, 25, 40 ,45 , 50 , 55 , 60 , 70 , 80 , 90 , 100
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT