Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1?

We recall that pH is defined by the formula $$\text{pH}= -\log_{10}[H^+]$$ where $$[H^+]$$ is the molar concentration of hydrogen ions in the final solution. For a strong acid-strong base mixture, the steps are: (i) calculate the initial moles of $$HCl$$ and $$NaOH$$, (ii) subtract to find the excess moles of the stronger component, (iii) divide by the total final volume to get $$[H^+]$$ (or $$[OH^-]$$), and (iv) use the definition of pH. We perform these steps for every option one by one.

Option A gives 55 mL of $$\dfrac{M}{10}$$ HCl and 45 mL of $$\dfrac{M}{10}$$ NaOH. First convert volumes into litres:

$$V_{\text{HCl}} = 55\text{ mL}=0.055\text{ L}, \quad V_{\text{NaOH}} = 45\text{ mL}=0.045\text{ L}$$

The molarity $$\dfrac{M}{10}$$ equals $$0.1\text{ M}$$. Using $$\text{moles}=M\times V$$,

$$n_{\text{HCl}} = 0.1 \times 0.055 = 0.0055$$

$$n_{\text{NaOH}} = 0.1 \times 0.045 = 0.0045$$

Excess moles of acid:

$$n_{H^+(\text{excess})}=0.0055-0.0045=0.0010$$

Total volume after mixing:

$$V_{\text{total}} = 0.055+0.045 = 0.100\text{ L}$$

Hydrogen-ion concentration:

$$[H^+]=\dfrac{0.0010}{0.100}=0.010\text{ M}$$

pH obtained:

$$\text{pH} = -\log_{10}(0.010)=2$$

We need pH 1, so Option A is not suitable.

Option B offers 75 mL of $$\dfrac{M}{5}$$ HCl and 25 mL of $$\dfrac{M}{5}$$ NaOH. Here $$\dfrac{M}{5}=0.2\text{ M}$$. Converting volumes:

$$V_{\text{HCl}} = 75\text{ mL}=0.075\text{ L}, \quad V_{\text{NaOH}} = 25\text{ mL}=0.025\text{ L}$$

Moles present:

$$n_{\text{HCl}} = 0.2 \times 0.075 = 0.015$$

$$n_{\text{NaOH}} = 0.2 \times 0.025 = 0.005$$

Excess acid moles:

$$n_{H^+(\text{excess})}=0.015-0.005 = 0.010$$

Total volume after mixing:

$$V_{\text{total}} = 0.075+0.025 = 0.100\text{ L}$$

Therefore the hydrogen-ion concentration is

$$[H^+]=\dfrac{0.010}{0.100}=0.10\text{ M}$$

Using the pH definition:

$$\text{pH} = -\log_{10}(0.10)=1$$

This mixture indeed gives pH 1.

Option C mixes 100 mL of $$\dfrac{M}{10}$$ HCl with 100 mL of $$\dfrac{M}{10}$$ NaOH. For each, $$M=0.1\text{ M}$$ and $$V=0.100\text{ L}$$, thus

$$n_{\text{HCl}} = 0.1 \times 0.100 = 0.010, \quad n_{\text{NaOH}} = 0.010$$

Both moles are equal, so complete neutralisation occurs, leaving no excess $$H^+$$ or $$OH^-$$:

$$[H^+]=10^{-7}\text{ M (pure water)}, \quad \text{pH}=7$$

Hence Option C does not satisfy the requirement.

Option D supplies 60 mL of $$\dfrac{M}{10}$$ HCl and 40 mL of $$\dfrac{M}{10}$$ NaOH. Converting:

$$V_{\text{HCl}} = 0.060\text{ L}, \quad V_{\text{NaOH}} = 0.040\text{ L}, \quad M = 0.1\text{ M}$$

Moles:

$$n_{\text{HCl}} = 0.1 \times 0.060 = 0.006$$

$$n_{\text{NaOH}} = 0.1 \times 0.040 = 0.004$$

Excess acid moles:

$$n_{H^+(\text{excess})}=0.006-0.004=0.002$$

Total volume:

$$V_{\text{total}} = 0.060+0.040=0.100\text{ L}$$

Hydrogen-ion concentration:

$$[H^+]=\dfrac{0.002}{0.100}=0.020\text{ M}$$

pH value:

$$\text{pH}=-\log_{10}(0.020)\approx 1.70$$

So Option D is also unsuitable.

Only Option B yields the required pH of 1.

Hence, the correct answer is Option B.