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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Chemistry
Chapter
Ionic Equilibrium

JIPMER 2019 · Chemistry · Ionic Equilibrium

If 50 ml of 0.1 HBr is mixed with 50 ml 0.2 M NaOH , find pH of...

If 50 ml of 0.1 HBr is mixed with 50 ml 0.2 M NaOH , find pH of resulting mixture

Options

  1. A2.7
  2. B12.7
  3. C10.7
  4. D1.3

Correct answer

B. 12.7

Step-by-step solution

NMix VMix =N1V1 (Base) −N2V2 (Acid) …(i)\mathrm{N}_{\text{Mix }} \mathrm{V}_{\text{Mix }}=\underset{\text{ (Base) }}{\mathrm{N}_1 \mathrm{ V}_1}-\underset{\text{ (Acid) }}{\mathrm{N}_2 \mathrm{ V}_2}\ldots(i) Given, M1=0.2M,N1=0.2NM2=0.1M,N2=0.1NV1=50ml,V2=50ml,Vmix =100ml\begin{aligned} \mathrm{M}_1 & =0.2 \mathrm{M}, \mathrm{N}_1=0.2 \mathrm{ N} \\ \mathrm{M}_2 & =0.1 \mathrm{M}, \mathrm{N}_2=0.1 \mathrm{ N} \\ \mathrm{ V}_1 & =50 \mathrm{ml}, \\ \mathrm{V}_2 & =50 \mathrm{ml}, \mathrm{V}_{\text{mix }}=100 \mathrm{ml} \end{aligned} On substituting the given values is equation. (i), We get Nmix ×100=50×0.2−50×0.1Nmix =5100=5×10−2∴[OH−]=5×10−2\begin{aligned} \mathrm{N}_{\text{mix }} \times 100 & =50 \times 0.2-50 \times 0.1 \\ \mathrm{ N}_{\text{mix }} & =\frac{5}{100}=5 \times 10^{-2} \\ \therefore \quad\left[\mathrm{OH}^{-}\right] & =5 \times 10^{-2} \end{aligned} Now, POH=−log⁡[OH−]=−log⁡[5×10−2]=2−log⁡5=1.3\begin{aligned} { }_{\mathrm{P}} \mathrm{OH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log \left[5 \times 10^{-2}\right] \\ & =2-\log 5=1.3 \end{aligned} Also, pH+pOH=14pH=14−1.3=12.7\mathrm{pH}+\mathrm{pOH}=14 \mathrm{pH}=14-1.3=12.7
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