KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 59 of 120

Marks: +1, -0

The pH of 0.1 M acetic acid solution is closest to [Dissociation constant of acid Ka = 1.8 × 10

^{–5}

]

Solution:

[H^{+ }] =√{ K_a .c}= √{1.8 ×10^{ –5}× 0.1 }

= √1.8 ×10^{–3}

pH = 3 – log (1.34) = 2.87

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