How do you calculate the pH at the equivalence point for the titration of 0.140 M methylamine ( $CH_3NH_2$ ) with 0.140 M $HCI$ ? The $K_b$ of methylamine is $5.0x10^-4$ ?

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How do you calculate the pH at the equivalence point for the titration of 0.140 M methylamine ( $CH_3NH_2$ ) with 0.140 M $HCI$ ? The $K_b$ of methylamine is $5.0x10^-4$ ?

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Answer 1 · 2017-09-03T05:35:53

pH = 8,57

Explanation:

at the equivalent point you have $CH_3NH_3Cl$ , and having unitet the two solutions you have a concentration of 0,07 M, The pH is given by
$[OH^-]=sqrt(Cs(Kw)/(Ka)) =sqrt (0,07 xx 10^(-14)/(5 xx 10^(-5))) = 3,7 xx 10^(-6)$
pOH= 5.43; pH = 8,57

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How do you calculate the pH at the equivalence point for the titration of 0.140 M methylamine ( $CH_3NH_2$ ) with 0.140 M $HCI$ ? The $K_b$ of methylamine is $5.0x10^-4$ ?

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Explanation:

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How do you calculate the pH at the equivalence point for the titration of 0.140 M methylamine (CH_3NH_2CH_3NH_2) with 0.140 M HCIHCI? The K_bK_b of methylamine is 5.0x10^-45.0x10^-4?

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Explanation:

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How do you calculate the pH at the equivalence point for the titration of 0.140 M methylamine ( $CH_3NH_2$ ) with 0.140 M $HCI$ ? The $K_b$ of methylamine is $5.0x10^-4$ ?