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NEET CHEMISTRYEasy

4 gm of NaOH is dissolved in 1000 ml of water. The H+\text{H}^+ ion concentration will be:

A

10−1 M10^{-1} \text{ M}

B

10−13 M10^{-13} \text{ M}

C

10−4 M10^{-4} \text{ M}

D

10−10 M10^{-10} \text{ M}

Step-by-Step Solution

First, we calculate the molarity of the NaOH\text{NaOH} solution. Molar mass of NaOH=23+16+1=40 g/mol\text{NaOH} = 23 + 16 + 1 = 40 \text{ g/mol}. Number of moles of NaOH=Given massMolar mass=4 g40 g/mol=0.1 mol\text{NaOH} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol}. Volume of solution =1000 ml=1 L= 1000 \text{ ml} = 1 \text{ L}. Molarity (MM) =Moles of soluteVolume of solution in L=0.11=0.1 M= \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.1}{1} = 0.1 \text{ M}. Since NaOH\text{NaOH} is a strong base, it completely dissociates into Na+\text{Na}^+ and OH−\text{OH}^- ions in aqueous solution . Therefore, the concentration of hydroxyl ions, [OH−]=0.1 M=10−1 M[\text{OH}^-] = 0.1 \text{ M} = 10^{-1} \text{ M}. We know the ionic product of water, Kw=[H+][OH−]=10−14K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} at 298 K298 \text{ K} . Thus, [H+]=Kw[OH−]=10−1410−1=10−13 M[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-1}} = 10^{-13} \text{ M}. Hence, the H+\text{H}^+ ion concentration is 10−13 M10^{-13} \text{ M}.

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