Calculate the pH of the resulting solution obtained by mixing 60\ mL0.1\ M HCl and 40\ mL0.1\ M NaOH.

Correct answer: B. 1.699 Millimoles of HCl = 60 × 0.1 = 6\ mmol Millimoles of NaOH = 40 × 0.1 = 4\ mmol Since HCl and NaOH react in a 1:1 ratio, the remaining millimoles of HCl = 6 - 4 = 2\ mmol . Total volume of the mixture = 60 + 40 = 100\ mL . Concentration of ^ + ions in the resulting solution is given by: [ ^ + ] = 2\ mmol 100\ mL = 0.02\ M pH = - [ ^ + ] = - (2 × 10^ -2 ) pH = 2 - 2 = 2 - 0.301 = 1.699 Answer: 1.699

MHT CET 202621 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 21 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Ionic Equilibrium

MHT CET 2026 · 21 April · Evening Shift · Chemistry · Ionic Equilibrium

Calculate the pH of the resulting solution obtained by mixing 60\...

Calculate the pH of the resulting solution obtained by mixing

60\ \text{mL}0.1\ \text{M}

HCl and

40\ \text{mL}0.1\ \text{M}

NaOH.

Options

A $2.301$
B $1.699$
C $2.699$
D $3.301$

Correct answer

B. $1.699$

Step-by-step solution

Millimoles of

\text{HCl} = 60 \times 0.1 = 6\ \text{mmol}

Millimoles of

\text{NaOH} = 40 \times 0.1 = 4\ \text{mmol}

Since

\text{HCl}

and

\text{NaOH}

react in a

1:1

ratio, the remaining millimoles of

\text{HCl} = 6 - 4 = 2\ \text{mmol}

.

Total volume of the mixture

= 60 + 40 = 100\ \text{mL}

.

Concentration of

\textH^{+}

ions in the resulting solution is given by:

[\textH^{+}] = \dfrac{2\ \text{mmol}}{100\ \text{mL}} = 0.02\ \text{M}

\text{pH} = -\log[\textH^{+}] = -\log(2 \times 10^{-2})

\text{pH} = 2 - \log 2 = 2 - 0.301 = 1.699

Answer:

1.699

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