Question

Tartaric acid is a diprotic acid we can represent as H₂T. Suppose a 100mL solution, which contains 0.1000M T^2- is titrated with 0.2000M HCl. K_a1=6.21x10^-5 and K_a2 = 2.22x10^-8.

a) What is the initial pH of the T^2- above? (before any HCl is added)

b) What is the pH after the addition of 100mL of 0.2000M HCl to the T^2- above?

Please show all work!

Answer 1

a Ans.

The dissociation of tartaric occurs in two steps.

Consider the first step,

The ICE table is

	H2T (aq)	H+(aq)	HT-(aq)
I	0.1000M	0	0
C	-x	+x	+x
E	0.1000-x	x	x

The expression for K_a1 is

(The value of 0.1000-x is approximated to 0.1000 since the value of x is very less. You can check if the approximation is valid or not)

To check if the approximation is valid or not, calculate the percentage of x with respect to the initial tartaric concentration (because x is dissociated from tartaric). If the percentage is less than 5%, then the approximation is valid.

% which is less than 5%.

So, the approximation is valid.

Now, consider the second step

The ICE table is

	HT- (aq)	H+ (aq)	T2- (aq)
I	2.49 x 10-3M	2.49 x 10-3M	0
C	-y	+y	+y
E	2.49 x 10-3-y	2.49 x 10-3+y	y

SInce the dissociation constant is very low, the value of y in (2.49 x 10^-3 + y) and (2.49 x 10^-3 - y) is neglected. This approximation can be checked.

or

To check if the approximation is valid or not,

% which is less than 5%.

So, the approximation is valid.

The total [H⁺] is given by

pH is given by

Therefore, pH of the tartaric acid solution before the addition of HCl is 2.604