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Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
a. 60 mL $\frac{M}{10}$ HCl + 40 mL $\frac{M}{10}$ NaOH;
b. 55 mL $\frac{M}{10}$ HCl + 45 mL $\frac{M}{10}$ NaOH;
c. 75 mL $\frac{M}{5}$ HCl + 25 mL $\frac{M}{5}$ NaOH;
d. 100 mL $\frac{M}{10}$ HCl + 100 mL $\frac{M}{10}$ NaOH.
pH of which one of them will be equal to 1?
a. 60 mL $\frac{M}{10}$ HCl + 40 mL $\frac{M}{10}$ NaOH;
b. 55 mL $\frac{M}{10}$ HCl + 45 mL $\frac{M}{10}$ NaOH;
c. 75 mL $\frac{M}{5}$ HCl + 25 mL $\frac{M}{5}$ NaOH;
d. 100 mL $\frac{M}{10}$ HCl + 100 mL $\frac{M}{10}$ NaOH.
pH of which one of them will be equal to 1?
Explanation
Only mixture c leaves $[H^+] = 0.1$ M.
Detailed Solution
For mixture c:
Meq of HCl $= 75\times\frac{1}{5}\times1 = 15$
Meq of NaOH $= 25\times\frac{1}{5}\times1 = 5$
Meq of HCl in resulting solution = 10
Molarity of $[H^+]$ in resulting mixture $= \frac{10}{100} = \frac{1}{10}$
$pH = -\log[H^+] = -\log\left(\frac{1}{10}\right) = 1.0$
So mixture c has pH = 1.