$(a)\;\;60\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+40 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
$(b)\;\;55\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+45 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
$(c)\;\;75\; \mathrm{mL} \frac{\mathrm{M}}{5}\; \mathrm{HCl}+25 \;\mathrm{mL} \frac{\mathrm{M}}{5} \;\mathrm{NaOH}$
$(d)\;\;100\; \mathrm{mL} \frac{\mathrm{M}}{10}\; \mathrm{HCl}+100 \;\mathrm{mL} \frac{\mathrm{M}}{10} \;\mathrm{NaOH}$
તેઓ પૈકી કોની $pH$ $1$ ને સમાન થશે ?
So acid is left at the end of reaction
\(\mathrm{N}_{\text {final solution }}=\left[\mathrm{H}^{+}\right] =\frac{\mathrm{N}_{1} \mathrm{V}_{1}-\mathrm{N}_{2} \mathrm{V}_{2}}{\mathrm{V}_{1}+\mathrm{V}_{2}}\)
\(=\frac{\frac{1}{5} \times 75-\frac{1}{5} \times 25}{75+25}\)
\(=\frac{1}{10}=0.1\)
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] =1\)
$A = NH_4Cl$; $ B = CH_3COONa$; $ C = NH_4OH$; $D = CH_3COOH$