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NEET CHEMISTRYMedium

The hydrogen ion concentration of a 10−8 M10^{-8}\text{ M} HCl aqueous solution at 298 K (Kw=10−14K_w = 10^{-14}) is:

A

1.0×10−6 M1.0 \times 10^{-6}\text{ M}

B

1.0525×10−7 M1.0525 \times 10^{-7}\text{ M}

C

9.525×10−8 M9.525 \times 10^{-8}\text{ M}

D

1.0×10−8 M1.0 \times 10^{-8}\text{ M}

Step-by-Step Solution

For very dilute acidic solutions like 10−8 M10^{-8}\text{ M} HCl, the contribution of H+H^+ from the auto-ionization of water cannot be neglected. Let the [H+][H^+] contributed by the dissociation of water be xx. Since water produces H+H^+ and OH−OH^- in equal amounts, [OH−]=x[OH^-] = x. The total concentration of hydrogen ions is the sum of those from HCl and water: Total [H+]=[H+]HCl+[H+]water=10−8+x[H^+] = [H^+]_{\text{HCl}} + [H^+]_{\text{water}} = 10^{-8} + x We know the ionic product of water is Kw=[H+][OH−]=10−14K_w = [H^+][OH^-] = 10^{-14} at 298 K. Substituting the values: (10−8+x)(x)=10−14(10^{-8} + x)(x) = 10^{-14} x2+10−8x−10−14=0x^2 + 10^{-8}x - 10^{-14} = 0 Solving this quadratic equation for xx: x=−10−8+(10−8)2−4(1)(−10−14)2x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2} x=−10−8+10−16+400×10−162x = \frac{-10^{-8} + \sqrt{10^{-16} + 400 \times 10^{-16}}}{2} x=−10−8+401×10−162x = \frac{-10^{-8} + \sqrt{401 \times 10^{-16}}}{2} x=−10−8+20.025×10−82=9.5125×10−8 Mx = \frac{-10^{-8} + 20.025 \times 10^{-8}}{2} = 9.5125 \times 10^{-8}\text{ M} Now, calculating the total [H+][H^+]: Total [H+]=10−8+9.5125×10−8=10.5125×10−8 M≈1.0525×10−7 M[H^+] = 10^{-8} + 9.5125 \times 10^{-8} = 10.5125 \times 10^{-8}\text{ M} \approx 1.0525 \times 10^{-7}\text{ M}.

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