Question

50 ml of 1 M CH3COOH solution, when added to 50 ml of 0.5 M NaoH
gives a solution with a pH value X. Find the value of 'X. If pka of
acetic acid is 4.8.
​

Answer 1

Answer:

Volume of CH3COOH = 50 ml

Molarity of CH3COOH = 1 M

Volume of NAOH = 50 ml

Molarity of NaOH = 0.5 M

pKa of CH3COOH = 4.8

The reaction is :

CH subscript 3 COOH space space plus space NaOH space rightwards arrow space CH subscript 3 COONa space plus space straight H subscript 2 straight O

Moles space of space acetic space acid space added space colon

straight c space equals straight n over straight V

straight n equals space straight c. straight V

straight n subscript open parentheses acetic space acid close parentheses end subscript space equals space 1 cross times 50 cross times 10 to the power of negative 3 end exponent space

space space space space space space space space space space space space space space space space space equals space 0.05 space mol

straight n subscript open parentheses NaOH close parentheses end subscript space equals space 0.5 space cross times 50 cross times 10 to the power of negative 3 end exponent

space space space space space space space space space space space space equals 0.025 space mol

Moles space of space acetic space acid space remained space equals 0.05 minus 0.025

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.025 space moles

It space forms space an space acidic space buffer space solution space as comma space both space weak space acid space and space its space salt space are space present space in space the space solution.

Total space volume space equals space 50 space plus space 50

space space space space space space space space space space space space space space space space space space space space equals space 100 space ml

space space space space space space space space space space space space space space space space space space space space space space space space space space space

space space space space space space space space space space space space space space space space space space space equals space 0.1 space litre

space open square brackets acid close square brackets space equals fraction numerator moles over denominator Volume space in space litre end fraction

space space space space space space space space space space equals fraction numerator 0.05 over denominator 0.1 end fraction

space space space space space space space space space space space equals 0.5 space straight M

open square brackets Salt space close square brackets equals fraction numerator 0.025 over denominator 0.1 end fraction

space space space space space space space space space space space space space equals space 0.25 space straight M

We space have comma

pH space equals pK subscript straight a space plus space log open parentheses fraction numerator open square brackets salt close square brackets over denominator open square brackets acid close square brackets end fraction close parentheses

space space space space space space equals 4.8 plus log open parentheses fraction numerator 0.5 over denominator 0.25 end fraction close parentheses

space space space space space space equals space 4.8 space plus space 0.3010

space pH space equals 5.1

Explanation:

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