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
1 answer:
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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