Calculate the pH of 1.0 times 10^{-8} M NaO | vedclass

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(B) For a strong base like $NaOH$,the concentration of $OH^-$ ions is $1.0 	imes 10^{-8} \ M$. However,we must also consider the contribution of $OH^

Vedclass · Accepted Answer

$7.02$

Vedclass · Answer

(B) For a strong base like $NaOH$,the concentration of $OH^-$ ions is $1.0 	imes 10^{-8} \ M$. However,we must also consider the contribution of $OH^-$ ions from the auto-ionization of water,which is $1.0 	imes 10^{-7} \ M$. Total $[OH^-] = 1.0 	imes 10^{-8} + 1.0 	imes 10^{-7} = 1.1 	imes 10^{-7} \ M$. $pOH = -\log[OH^-] = -\log(1.1 	imes 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.041 = 6.959$. Since $pH + pOH = 14$,$pH = 14 - 6.959 = 7.041$. Rounding to the nearest provided option,the value is $7.02$.

Calculate the $pH$ of $1.0 \times 10^{-8} \ M \ NaOH$ solution.

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