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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

30 ml of 0.2 M NaOH is added with 50 ml 0.2 M CH _ 3 COOH solution....

30 ml of 0.2 M NaOH is added with 50 ml 0.2 M CH3COOH{CH}_{3}COOH solution. The extra volume of 0.2 M NaOHNaOH required to make the pH of the solution 5.00 is 10x\dfrac{10}{x} .The value of xx is. The ionisation constant of CH3COOH=2×10−5{CH}_{3}COOH = 2 \times 1{0}^{- 5} .30 ml of 0.2 M NaOH is added with 50 ml 0.2 M CH _ 3 COOH solution. The extra volume of 0.2 M NaOH required to make the

Correct answer

3

Step-by-step solution

Milli.moles of NaOH=30×0.2=6NaOH = 30 \times 0.2 = 6

m.m of CH3COOH=50×0.2=10C{H}_{3}COOH = 50 \times 0.2 = 10

Now:

30 ml of 0.2 M NaOH is added with 50 ml 0.2 M CH _ 3 COOH solution. The extra volume of 0.2 M NaOH required to make the

pH=−log⁡(2×10−5)+log64pH = - \log (2 \times 1{0}^{- 5} ) + log \dfrac{6}{4}

=4.87= 4.87

Suppose 'v'mL of NaOHNaOH is added then

m.mol of CH3COONa=(6+v×0.2){CH}_{3}COONa = \left(6 + v \times 0.2\right)

m.mol of CH3COOH=(4−v×0.2){CH}_{3}COOH = \left(4 - v \times 0.2\right)

pH=pKa+log⁡[SaltAcid]\text{pH} = {\text{pK}}_{\text{a}} + \log \left[\dfrac{\text{Salt}}{\text{Acid}}\right]
5=−log⁡(2×10−5)+log⁡(6+0.2v4−0.2v)5 = - \log (2 \times 1{0}^{- 5} ) + \log \left(\dfrac{6 + 0.2v}{4 - 0.2v}\right)

0.3010=log⁡(6+0.2v4−0.2v)0.3010 = \log \left(\dfrac{6 + 0.2v}{4 - 0.2v}\right)

So \left(\dfrac{6 + 0.2v}{4 - 0.2v}\right) = 2 ⇒ v = 3.33mL = \dfrac{10}{3}mL<br/><br/>So<br /><br />Sox = 3
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