Equal volumes of two solutions A and B of a s | vedclass

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(C) For solution $A$: $pH=6.0$,so $[H^{+}]_A = 10^{-6} \ M$. For solution $B$: $pH=4.0$,so $[H^{+}]_B = 10^{-4} \ M$. Let the volume of each solution

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between $4$ and $5$

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(C) For solution $A$: $pH=6.0$,so $[H^{+}]_A = 10^{-6} \ M$. For solution $B$: $pH=4.0$,so $[H^{+}]_B = 10^{-4} \ M$. Let the volume of each solution be $1 \ L$. Total moles of $H^{+}$ = $(1 \ L 	imes 10^{-6} \ M) + (1 \ L 	imes 10^{-4} \ M) = 10^{-6} + 10^{-4} = 1.01 	imes 10^{-4} \ 	ext{moles}$. Total volume of the mixture = $1 \ L + 1 \ L = 2 \ L$. Resultant $[H^{+}] = \frac{1.01 	imes 10^{-4} \ 	ext{moles}}{2 \ L} = 5.05 	imes 10^{-5} \ M$. $pH = -\log(5.05 	imes 10^{-5}) = 5 - \log(5.05) \approx 5 - 0.703 = 4.297$. Since $4.297$ lies between $4$ and $5$,the correct range is between $4$ and $5$.

Equal volumes of two solutions $A$ and $B$ of a strong acid having $pH=6.0$ and $pH=4.0$ respectively are mixed together to form a new solution. The $pH$ of the new solution will be in the range

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