Test Index

BITSAT 2022 Paper

Section: Chemistry

Question : 48 of 130

Marks: +1, -0

100\ \mathrm{mL}

2\ \mathrm{M}

\text{acid} ( \mathrm{p} K_a=3.74)

\mathrm{NaOH}

\mathrm{pH}

Solution:

Sodium formate is present at the equivalence point. It is the salt of weak acid + strong base. So, final solution will be basic in nature.

As we know

\mathrm{pH} = 7 + \frac{\mathrm{pK_a}}{2} + \frac{\log C}{2}

where,

C

is concentration of salt.

Total volume of solution

= 100 + 100 = 200\ \mathrm{mL}

Concentration of salt

(C)=2 \times \frac{100}{200}=1\mathrm{M}

\mathrm{pH}=7 + \frac{3.74}{2} + \frac{\log(1)}{2}

\mathrm{pH}=7+1.87+0

\mathrm{pH}=8.87

Go to Question:

Prev Question

Next Question