Which $pH$ value is higher in the following pairs?
$(a)$ $0.1 \ M \ HCl$ and $0.1 \ M \ NaOH$
$(b)$ $0.1 \ M \ HCl$ and $0.01 \ M \ HCl$
$(c)$ $0.1 \ M \ NaOH$ and $0.01 \ M \ NaOH$
$(a)$ $0.1 \ M \ HCl$ and $0.1 \ M \ NaOH$
$(b)$ $0.1 \ M \ HCl$ and $0.01 \ M \ HCl$
$(c)$ $0.1 \ M \ NaOH$ and $0.01 \ M \ NaOH$
(N/A) For $0.1 \ M \ HCl$,$pH = -\log(0.1) = 1$. For $0.1 \ M \ NaOH$,$pOH = -\log(0.1) = 1$,so $pH = 14 - 1 = 13$. Thus,$0.1 \ M \ NaOH$ has a higher $pH$.
$(b)$ For $0.1 \ M \ HCl$,$pH = 1$. For $0.01 \ M \ HCl$,$pH = -\log(0.01) = 2$. Thus,$0.01 \ M \ HCl$ has a higher $pH$.
$(c)$ For $0.1 \ M \ NaOH$,$pOH = 1$,so $pH = 13$. For $0.01 \ M \ NaOH$,$pOH = -\log(0.01) = 2$,so $pH = 14 - 2 = 12$. Thus,$0.1 \ M \ NaOH$ has a higher $pH$.
$(b)$ For $0.1 \ M \ HCl$,$pH = 1$. For $0.01 \ M \ HCl$,$pH = -\log(0.01) = 2$. Thus,$0.01 \ M \ HCl$ has a higher $pH$.
$(c)$ For $0.1 \ M \ NaOH$,$pOH = 1$,so $pH = 13$. For $0.01 \ M \ NaOH$,$pOH = -\log(0.01) = 2$,so $pH = 14 - 2 = 12$. Thus,$0.1 \ M \ NaOH$ has a higher $pH$.
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