Question

A solution is made that is 0.400 M H3BO3(Ka= 5.4×10−10) and 0.125 M HCNO (Ka= 2×10−4).

Which answer approximates the pH of the solution:

A) 0.28

B) 0.90

C) 2.05

D) 2.30

E) 4.83

Answer 1

As we have

Ka (HCNO) >> Ka(H3BO3)

So disassociation of HCNO will dominate this mixture

HCNO -> H⁺ + CNO^{​​​​ -}

t= 0 0.125 0 0

Disassociation -x +x + x

t= t(eq) 0.125 - x x x

Ka = 2 × 10 ^-4 = ( x ² / 0.125 - x ) neglect x as x<<0.125

Now x^{​​​​​​ 2} = 0.125 × 2× 10 ^-4

x = 0.005 --- (1)

{ Now disassociation

= (0.005/0.125) × 100

= 4 %

it is less than 5% . If it increases from 5 % we can not neglect "x" in our calculation.

}

Now for H3BO3

H3BO3 -> H ⁺ + H2BO3 ^-

t= 0 0.400 0.005 0

Disassociation -x + x +x

Equilibrium 0.400- x 0.005 + x x

Now

Ka = 5.4 × 10 ^-10 = (0.005 + x )(x) / (0.400 - x)

On neglecting x

5.4 × 10 ^-10 =( 0.005)( x) / (0.400)

x = 4.32 × 10 ^-8 ----- (2)

( 2) << (1) so (2) can be neglected

Now calculating

pH = - log([H+])

pH = - log (0.005)

pH = 2.30 (D)