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Question: Ph of 10^-7M naoh...

Ph of 10^-7M naoh

Answer

7.21

Explanation

Solution

To determine the pH of a 10−710^{-7} M NaOH solution, we must consider the autoionization of water, as the concentration of the base is comparable to the concentration of H+H^+ and OH−OH^- ions in pure water (10−710^{-7} M at 25°C).

  1. Identify the sources of OH−OH^- ions:

    • From NaOH (a strong base, fully dissociates): [OH−]NaOH=10−7[OH^-]_{NaOH} = 10^{-7} M.
    • From water autoionization: H2O(l)⇌H+(aq)+OH−(aq)H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq). Let the concentration of H+H^+ ions produced by water be xx. Due to stoichiometry, the concentration of OH−OH^- ions produced by water will also be xx.

  2. Express total ion concentrations:

    • Total [H+][H^+] in solution: [H+]total=x[H^+]_{total} = x
    • Total [OH−][OH^-] in solution: [OH−]total=[OH−]NaOH+[OH−]water=10−7+x[OH^-]_{total} = [OH^-]_{NaOH} + [OH^-]_{water} = 10^{-7} + x

  3. Apply the ion product of water (KwK_w):

    At 25°C, Kw=[H+]total[OH−]total=10−14K_w = [H^+]_{total}[OH^-]_{total} = 10^{-14}. Substitute the expressions from step 2: x(10−7+x)=10−14x(10^{-7} + x) = 10^{-14}

  4. Formulate and solve the quadratic equation:

    10−7x+x2=10−1410^{-7}x + x^2 = 10^{-14} x2+10−7x−10−14=0x^2 + 10^{-7}x - 10^{-14} = 0

    Using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}: Here, a=1a=1, b=10−7b=10^{-7}, c=−10−14c=-10^{-14}. x=−10−7±(10−7)2−4(1)(−10−14)2(1)x = \frac{-10^{-7} \pm \sqrt{(10^{-7})^2 - 4(1)(-10^{-14})}}{2(1)} x=−10−7±10−14+4×10−142x = \frac{-10^{-7} \pm \sqrt{10^{-14} + 4 \times 10^{-14}}}{2} x=−10−7±5×10−142x = \frac{-10^{-7} \pm \sqrt{5 \times 10^{-14}}}{2} x=−10−7±5×10−72x = \frac{-10^{-7} \pm \sqrt{5} \times 10^{-7}}{2}

    Since xx represents a concentration, it must be positive. We take the positive root. 5≈2.236\sqrt{5} \approx 2.236 x=−10−7+2.236×10−72x = \frac{-10^{-7} + 2.236 \times 10^{-7}}{2} x=(2.236−1)×10−72x = \frac{(2.236 - 1) \times 10^{-7}}{2} x=1.236×10−72x = \frac{1.236 \times 10^{-7}}{2} x=0.618×10−7x = 0.618 \times 10^{-7} M So, [H+]=6.18×10−8[H^+] = 6.18 \times 10^{-8} M.

  5. Calculate pH:

    pH=−log⁡[H+]pH = -\log[H^+] pH=−log⁡(6.18×10−8)pH = -\log(6.18 \times 10^{-8}) pH=−(log⁡6.18+log⁡10−8)pH = -(\log 6.18 + \log 10^{-8}) pH=−(log⁡6.18−8)pH = -(\log 6.18 - 8) pH=8−log⁡6.18pH = 8 - \log 6.18

    Using log⁡6.18≈0.791\log 6.18 \approx 0.791: pH=8−0.791=7.209pH = 8 - 0.791 = 7.209

Alternatively, calculate pOH first: [OH−]total=10−7+x=10−7+0.618×10−7=(1+0.618)×10−7=1.618×10−7[OH^-]_{total} = 10^{-7} + x = 10^{-7} + 0.618 \times 10^{-7} = (1 + 0.618) \times 10^{-7} = 1.618 \times 10^{-7} M. pOH=−log⁡[OH−]totalpOH = -\log[OH^-]_{total} pOH=−log⁡(1.618×10−7)pOH = -\log(1.618 \times 10^{-7}) pOH=−(log⁡1.618+log⁡10−7)pOH = -(\log 1.618 + \log 10^{-7}) pOH=−(log⁡1.618−7)pOH = -(\log 1.618 - 7) pOH=7−log⁡1.618pOH = 7 - \log 1.618 Using log⁡1.618≈0.2089\log 1.618 \approx 0.2089: pOH=7−0.2089=6.7911pOH = 7 - 0.2089 = 6.7911

Finally, pH=14−pOHpH = 14 - pOH pH=14−6.7911=7.2089pH = 14 - 6.7911 = 7.2089

Rounding to two decimal places, the pH is 7.21.

The final answer is 7.21\boxed{\text{7.21}}.