What is the pH of a 2.6 times 10^{-8} M H^{ | vedclass

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(B) The formula for $pH$ is $pH = -\log [H^{+}]$.Since the concentration of $H^{+}$ is very low $(< 10^{-7} \ M)$,the contribution of $H^{+}$ from the

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$6.9$

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(B) The formula for $pH$ is $pH = -\log [H^{+}]$.Since the concentration of $H^{+}$ is very low $($[H^{+}]_{total} = [H^{+}]_{acid} + [H^{+}]_{water} = 2.6 	imes 10^{-8} + 1.0 	imes 10^{-7} \ M$.Converting to the same power of $10$: $[H^{+}]_{total} = 0.26 	imes 10^{-7} + 1.0 	imes 10^{-7} = 1.26 	imes 10^{-7} \ M$.Now,$pH = -\log(1.26 	imes 10^{-7}) = 7 - \log(1.26)$.Given $\log 2.6 = 0.4150$,we know $\log 1.26 = \log(2.6 / 2) = \log 2.6 - \log 2 = 0.4150 - 0.3010 \approx 0.114$.$pH = 7 - 0.114 = 6.886 \approx 6.9$.

What is the $pH$ of a $2.6 \times 10^{-8} \ M \ H^{+}$ ion solution?
$(\log 2.6 = 0.4150)$

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What is the $pH$ of a $2.6 \times 10^{-8} \ M \ H^{+}$ ion solution?$(\log 2.6 = 0.4150)$