Question

A 10.00 mL aliquot of a 0.02000 F neutral ethylenediamine (H2NCH2CH2NH2) solution is to be titrated.

The titrant HCL is 0.0500 F

(Ka value is 1.18*10^-10 Pka value is 9.92)

Question:

9. What is the pH after adding 8.00 mL of the titrant?

Answer 1

Ethylenediamine = H2NCH2CH2NH2) = 10.0mL of 0.02000F

number of moles of Ethylenediamine = 0.0200Fx0.010L= 0.0002 moles

HCl = 8.00Ml of 0.0500F

number o fmolesof HCl = 0.0500F x0.008L = 0.0004 moles

number of moles of HCl is greatger than the number of moles of Base

remaining number of moles of HCl = 0.0004 - 0.0002 = 0.0002 moles

Total volume = 10.00+8.00=18.00mL= 0.018L

[H+] = number of moles/volume = 0.0002/0.018 =0.011M

[H+] = 0.011M

-log[H+] = -log(0.011)

PH= 1.958

PH= 1.96

{ formality is equal to molarity)