Calculate the \( \mathrm{pH} \) of the resultant mixtures: a) \( 10...

Calculate the \( \mathrm{pH} \) of the resultant mixtures:
a) \( 10 \mathrm{~mL} \) of \( 0.2 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}+25 \mathrm{~mL} \) of \( 0.1 \mathrm{M} \mathrm{HCl} \)
b) \( 10 \mathrm{~mL} \) of \( 0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL} \) of \( \left.0.01 \mathrm{M} \mathrm{Ca(OH}\right)_{2} \)
c) \( 10 \mathrm{~mL} \) of \( 0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL} \) of \( 0.1 \mathrm{M} \mathrm{KOH} \)
\( p O H=1.36 \quad \mathrm{pH}+\mathrm{pOH}=14 \)
\( p H=14-1.36=\frac{12.63}{\text { Ans }} \)
(b)
\[
\begin{array}{l}
n_{\mathrm{H}^{+}}=\frac{0.01 \times 10}{1000} \times 2=0.0002 \mathrm{~mol} \leftarrow \\
n_{\mathrm{OH}^{-}}=\frac{0.01 \times 10}{1000} \times 2=0.0002 \mathrm{~mol}
\end{array}
\]
Physics Wallah

[Anshsingh-ee]

great concept clarity sir thankyou very much

[kartiksingh]

Excellent video for examinees. Thanks sir 🙏

[shuklanaimish]

Cleared all my concepts for these type of questions

[sunghoniee]

In 4:58 how can they be neutral? It’s not the same number one of them is 0.002 and the other is 0.0002