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Question: The pH of $10^{-8}$ M HCl solution is...

The pH of 10−810^{-8} M HCl solution is

A

8

B

More than 8

Answer

Approximately 6.98, none of the provided options are correct.

Explanation

Solution

The pH of a solution is determined by the total concentration of hydrogen ions ([H+][H^+]). For very dilute acid solutions, the contribution of H+H^+ ions from the autoionization of water cannot be neglected.

  1. Concentration of HCl: The given concentration of HCl is 10−810^{-8} M. Since HCl is a strong acid, it completely dissociates in water, contributing 10−810^{-8} M H+H^+ ions. So, [H+]acid=10−8[H^+]_{acid} = 10^{-8} M.

  2. Autoionization of Water: Water undergoes autoionization: H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-. The ion product of water, KwK_w, is 1.0×10−141.0 \times 10^{-14} at 25°C, where Kw=[H+][OH−]K_w = [H^+][OH^-].

  3. Total Hydrogen Ion Concentration: Let the total hydrogen ion concentration be [H+]total[H^+]_{total} and the total hydroxide ion concentration be [OH−]total[OH^-]_{total}. The H+H^+ ions come from both HCl and water. The OH−OH^- ions come only from water's autoionization. Let xx be the concentration of OH−OH^- ions produced from water's autoionization. Then, the concentration of H+H^+ ions produced from water's autoionization is also xx. So, [OH−]total=x[OH^-]_{total} = x. And [H+]total=[H+]acid+[H+]water_from_autoionization=10−8+x[H^+]_{total} = [H^+]_{acid} + [H^+]_{water\_from\_autoionization} = 10^{-8} + x.

  4. Using the KwK_w expression: [H+]total[OH−]total=Kw[H^+]_{total} [OH^-]_{total} = K_w (10−8+x)(x)=1.0×10−14(10^{-8} + x)(x) = 1.0 \times 10^{-14} x2+10−8x−1.0×10−14=0x^2 + 10^{-8}x - 1.0 \times 10^{-14} = 0

  5. Solving the quadratic equation for xx: Using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}: x=−10−8±(10−8)2−4(1)(−1.0×10−14)2(1)x = \frac{-10^{-8} \pm \sqrt{(10^{-8})^2 - 4(1)(-1.0 \times 10^{-14})}}{2(1)} x=−10−8±10−16+4.0×10−142x = \frac{-10^{-8} \pm \sqrt{10^{-16} + 4.0 \times 10^{-14}}}{2} x=−10−8±0.01×10−14+4.0×10−142x = \frac{-10^{-8} \pm \sqrt{0.01 \times 10^{-14} + 4.0 \times 10^{-14}}}{2} x=−10−8±4.01×10−142x = \frac{-10^{-8} \pm \sqrt{4.01 \times 10^{-14}}}{2} x=−10−8±4.01×10−72x = \frac{-10^{-8} \pm \sqrt{4.01} \times 10^{-7}}{2} Since xx must be positive, we take the positive root: x=−10−8+2.0025×10−72x = \frac{-10^{-8} + 2.0025 \times 10^{-7}}{2} (approximately, as 4.01≈2.0025\sqrt{4.01} \approx 2.0025) x=−0.1×10−7+2.0025×10−72x = \frac{-0.1 \times 10^{-7} + 2.0025 \times 10^{-7}}{2} x=1.9025×10−72≈9.51×10−8x = \frac{1.9025 \times 10^{-7}}{2} \approx 9.51 \times 10^{-8} M

  6. Calculating Total [H+][H^+]: [H+]total=10−8+x=10−8+9.51×10−8[H^+]_{total} = 10^{-8} + x = 10^{-8} + 9.51 \times 10^{-8} [H+]total=(1+9.51)×10−8=10.51×10−8[H^+]_{total} = (1 + 9.51) \times 10^{-8} = 10.51 \times 10^{-8} [H+]total=1.051×10−7[H^+]_{total} = 1.051 \times 10^{-7} M

  7. Calculating pH: pH=−log⁡[H+]totalpH = -\log[H^+]_{total} pH=−log⁡(1.051×10−7)pH = -\log(1.051 \times 10^{-7}) pH=−(log⁡(1.051)+log⁡(10−7))pH = -(\log(1.051) + \log(10^{-7})) pH=−(log⁡(1.051)−7)pH = -(\log(1.051) - 7) pH=7−log⁡(1.051)pH = 7 - \log(1.051) Since log⁡(1.051)≈0.021\log(1.051) \approx 0.021, pH≈7−0.021=6.979pH \approx 7 - 0.021 = 6.979

The pH of the 10−810^{-8} M HCl solution is approximately 6.98. This value is slightly less than 7, which is expected for an acidic solution. Therefore, none of the given options are correct. The value 8 would be obtained if the contribution of water's autoionization were incorrectly neglected (pH=−log⁡(10−8)=8pH = -\log(10^{-8}) = 8), which is a common error for very dilute acid solutions.