Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. (NEET – 2018) i. 60 mL $\frac{M}{10}$ HCl + 40 mL $\frac{M}{10}$ NaOH ii. 55 mL $\frac{M}{10}$ HCl + 45 mL $\frac{M}{10}$ NaOH iii. 75 mL $\frac{M}{5}$ HCl + 25 mL $\frac{M}{5}$ NaOH iv. 100 mL $\frac{M}{10}$ HCl + 100 mL $\frac{M}{10}$ NaOH pH of which one of them will be equal to 1? a) iv b) i c) ii d) iii

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. (NE

Choose the Best Answer · Q2

\frac{M}{10}

a) iv

b) i

c) ii

d) iii

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Concept understanding — pH and pOH Scale

Acid/base solutions typically span the concentration range $10^{-1}$ to $10^{-7}$ M, an inconveniently wide span. Sorensen's logarithmic pH scale expresses this compactly: $pH=-\log_{10}[H_3O^+]$ , so $[H_3O^+]=10^{-pH}$ ; the analogous quantity for hydroxide is $pOH=-\log_{10}[OH^-]$ .

pH + pOH = pKw. Adding the two definitions gives $pH+pOH=-\log_{10}([H_3O^+][OH^-])=-\log_{10}K_w=pK_w$ . At $25^\circ C$ , $K_w=1\times10^{-14}$ , so $pK_w=14$ and $pH+pOH=14$ . In a neutral solution at $25^\circ C$ , both pH and pOH equal 7; acidic solutions have pH < 7, basic solutions have pH > 7.

Working with strong acids/bases: for a strong acid or base, dissociation is essentially complete, so $[H_3O^+]$ (or $[OH^-]$ ) can usually be read directly off the given concentration -- e.g. 0.001M HCl gives $pH=-\log_{10}(10^{-3})=3$ . When mixing a strong acid with a strong base, first find the moles of each, subtract to find which is in excess, then divide by the total volume to get the excess $[H^+]$ or $[OH^-]$ before taking $-\log_{10}$ .

When water's own ionisation cannot be ignored. If the acid/base concentration is below about $10^{-6}$ M, water's own $10^{-7}$ M contribution of $H_3O^+$ (or $OH^-$ ) is no longer negligible and must be added on: $[H_3O^+]=10^{-7}(\text{from water})+[H_3O^+](\text{from the acid})$ -- e.g. $10^{-7}$ M HCl gives $[H_3O^+]=10^{-7}+10^{-7}=2\times10^{-7}$ M and $pH=6.70$ , not the naive pH = 7.

Step 1. Convert each concentration: $\frac{M}{10}=0.1$ M, $\frac{M}{5}=0.2$ M. In each option, total volume after mixing is 100 mL.

Step 2 (option i). HCl = $60\times0.1=6$ mmol; NaOH = $40\times0.1=4$ mmol. Excess HCl = 2 mmol in 100 mL $\Rightarrow$ $[H^+]=0.02$ M $\Rightarrow pH=-\log(0.02)=1.70$ .

Step 3 (option ii). HCl = $55\times0.1=5.5$ mmol; NaOH = $45\times0.1=4.5$ mmol. Excess HCl = 1 mmol in 100 mL $\Rightarrow$ $[H^+]=0.01$ M $\Rightarrow pH=2$ .

Step 4 (option iii). HCl = $75\times0.2=15$ mmol; NaOH = $25\times0.2=5$ mmol. Excess HCl = 10 mmol in 100 mL $\Rightarrow$ $[H^+]=0.1$ M $\Rightarrow pH=-\log(0.1)=1$ . This matches.

Step 5 (option iv). HCl = $100\times0.1=10$ mmol = NaOH = $100\times0.1=10$ mmol $\Rightarrow$ exact neutralisation, pH = 7.

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(d) iii

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