Cyanic acid is a weak monoprotic acid. If the pH of 0.150 M cyanic acid is 2.32.Calculate Ka for cyanic acid.
pH=2.32pH=2.32pH=2.32
pH=−log([H3O+])pH=-log([H_3O^+])pH=−log([H3O+])
log([H3O+])=−pHlog([H_3O^+])=-pHlog([H3O+])=−pH
10log([H3O+])=10−pH10^{log([H_3O^+])}=10^{-pH}10log([H3O+])=10−pH
([H3O+])=10−pH([H_3O^+])=10^{-pH}([H3O+])=10−pH
([H3O+])=10−2.32=4.8×10−3M([H_3O^+])=10^{-2.32}=4.8×10^{-3}M([H3O+])=10−2.32=4.8×10−3M
[A−]=([H3O+])[A^-]=([H_3O^+])[A−]=([H3O+]) Produced in 1:1 mole
[A−]=4.8×10−3M[A^-]=4.8×10^{-3}M[A−]=4.8×10−3M
HA=0.150−[4.8×10−3]HA=0.150-[4.8×10^{-3}]HA=0.150−[4.8×10−3]
HA=0.1452MHA=0.1452MHA=0.1452M
Ka=[A−]×[H3O+][HA]K_a=\frac{[A^- ]× [H_3O^+]}{[HA]}Ka=[HA][A−]×[H3O+]
Ka=[4.8×10−3]20.1452K_a=\frac {[4.8×10^{-3}]^2}{0.1452}Ka=0.1452[4.8×10−3]2
Ka=1.6×10−4MK_a= 1.6 ×10^{-4}MKa=1.6×10−4M
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