If the ka of a monoprotic weak acid is 8.4 × 10-6, what is the ph of a 0.45 m solution of this acid

Consider the acid to be HA so the equation for acid will be:

HA+H₂O>H₃O⁺ +A⁻

Initial: 0.45 0 0 0

Final 0.45-x x x x

Kₐ= [H₃O⁺][A⁻]/[HA]

= x²/0.45-x

8.4 × 10⁻⁶= x²/0.45-x

3.78× 10⁻⁶-8.4 × 10⁻⁶x=x²

On solving equation,

x= 0.0018

[H₃O⁺]=[A⁻]=x=0.0018

pH= -log[H₃O⁺]

= -log[0.0018]

= 2.74

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