Question

A) calculate the ph of a 0.028 M HNO3 solution

B) Calculate the pH of a 0.028 M solution of benzoic acid (Ka=6.5x10^-5)

Answer 1

A- The given acid HNO3 is a strong acid. We know the strong acids dissociates 100% to H+ ion. Thus the pH of strong acid is calculated as

pH = -log [H+]

Now due to 100% dissociation i.e all the HNO3 will completely converted into H+, [H+] = [HNO3] =  0.028 M

Thus pH = -log [H+] =  -log [0.028] = 1.55

B-

For Benzoic acid (C₆H₅COOH), it is a weak acid. Thus it cant dissociates 100% to H+ ion. Hence some of the acid dissociate ti H+ and some remain as C₆H₅COOH.

Now the given Ka represents the dissociation constant for C₆H₅COOH

i.e C₆H₅COOH → H⁺ + C₆H₅COO^- This is the dissociation reaction of Benzoic acid

Here  C₆H₅COOH = the weak acid and C₆H₅COO^- = conjugate base

Here Ka = [H⁺][C₆H₅COO^-]/[C₆H₅COOH]

Now pH of weak acid is calculated by Henderson Hasselblach Equation-

pH = pKa + log [Conjugate base] / [weak acid] where pKa = -log Ka

Now given [C₆H₅COOH] = 0.028 M and Ka = 6.5x10^-5

Lets consider x M of acid is dissociate to C₆H₅COO^- . Then to find the concentration of acicd and conjugate base, we have to form the ICE table

	[C6H5COOH]	[H+]	[C6H5COO-]
Initial	0.028 M	0	0
Change	-x	+x	+x
Equilibrium	0.028 M -x	+x	+x

Now putting the values for

Ka = [H⁺][C₆H₅COO^-]/[C₆H₅COOH]

6.5x10^-5 = x * x / 0.028 M -x

6.5x10^-5 * (0.028 M -x) = x²

0.00000182‬ - 0.000065x = x²

Or x²‬ + 0.000065x - 0.00000182 = 0

Now solving this,

x = 0.0013

Thus at equilibrium, [C₆H₅COOH] = 0.028 M -x = 0.028 M - 0.0013 = 0.0267‬ M

And [C₆H₅COO^-] = x =  0.0013 M

Again pKa = -log Ka = -log (6.5x10^-5) = 5 - log 6.5 = 4.18

Now putting the values in the equation-

pH = pKa + log [Conjugate base] / [weak acid]

pH = 4.18 + log [0.0013] / [0.0267‬]

= 4.18 + log [0.048‬]

= 2.86