Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?

Correct answer: B. 75 ~mL M 5 HCl+25 ~mL M 5 NaOH 75 ~mL M 5 HCl+25 ~mL M 5 NaOH25 ~mL M 5 NaOH will neutralise 25 ~mL M 5 HCl75-25=50 ~mL M 5 HCl will remain. Total volume will be 75+25=100 ~mL50 ~mL M 5 HCl is diluted to 100 ~mL[H^ + ]=[HCl]= M 5 × (50)/(100)= M 10 pH=- _ 10 [H^ + ]=- _ 10 M 10 = 1

JEE Main 201815 AprilEvening ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Evening Shift
Mode: Online
Subject: Chemistry
Chapter: Ionic Equilibrium

JEE Main 2018 · 15 April · Evening Shift · Online · Chemistry · Ionic Equilibrium

Following four solutions are prepared by mixing different volumes...

Following four solutions are prepared by mixing different volumes of

\mathrm{NaOH}

and

\mathrm{HCl}

of different concentrations,

\mathrm{pH}

of which one of them will be equal to 1 ?

Options

A $55 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+45 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$
B $75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$
C $100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$
D $60 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+40 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$

Correct answer

B. $75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$

Step-by-step solution

75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}

will neutralise

25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}75-25=50 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}

will remain. Total volume will be

75+25=100 \mathrm{~mL}50 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}

is diluted to

100 \mathrm{~mL}\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=\frac{\mathrm{M}}{5} \times \frac{50}{100}=\frac{\mathrm{M}}{10}\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=-\log _{10} \frac{\mathrm{M}}{10}=

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