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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