A chemist has synthesized a monoprotic weak acid and wants to determine its Ka value. To do so, the chemist dissolves 0.002 moles of the acid in a solution with a total volume of 100.0 ml. He titrates the acid with 0.0500 M NaOH. After 20.0 mL of NaOH is added, the pH is 6.00. What is the Ka value for this acid? (show work below)

 

Suppose HA is a monoprotic weak acid.

The ionization equilibrium HA is written as:

HA(aq) ----> A^-(aq) + H⁺(aq)

Write Ka expression:

Ka = [H⁺] [A^-] /[ HA]

We can calculate Ka from the salt-to-acid ratio and the measured pH.

Calculate the salt-to-acid ratio:

Moles of acid HA = 0.002 moles

Moles of NaOH =0.0500 M * 20.0 mL *1L/1000mL =0.001 moles

The balanced equation for neutralization reaction is written as:

HA (aq) + NaOH(aq) -> H₂O(l) + NaA(aq)

Balanced equation shows the molar ratio between HA and NaOH is 1:1

0.001 moles of NaOH reacts with 0.001 moles of HA to produce 0.001moles of A^-

The amount of HA left =0.002moles - 0.001 moles = 0.001 moles

Total volume of solution = 100.0 ml + 20.0 ml = 120.00 ml = 0.120 L

[A^-] = 0.001moles/0.120 L =0.0083M

[HA] =0.001moles/0.120 L =0.0083M

 [A^-] /[HA] =0.0083 / 0.0083 =1

Calculate [H⁺] from pH:

The pH is 6.00

pH = - log [H⁺]

[H₃O⁺] = 10^-PH    

[H₃O⁺] = 10^-(6.00)

[H⁺]= 1.0 x 10^-6 M

Now plug the value of [A^-][H⁺] and[ HA] in Ka expression and calculate Ka monoprotic weak acid:

Ka = [H⁺] *[A^-] /[ HA]

Ka =1.0 x 10^-6*(1) =1.0 x 10^-6

Ka value for a monoprotic weak acid is 1.0 x 10^-6