Calculate the pH of abuffer solution that contains 0.25M HCNO and 0.10M NaCNO. The Ka of HCNO is 2.2 x 10^-4
How is this done?
a.) pH=3.51
b.) pH=3.81
c.) pH=4.06
d.) pH=3.26 * I know this is the answer but I don't understand how to get this.
Calculate the pH of abuffer solution that contains 0.25M HCNO and 0.10M NaCNO. The Ka of HCNO is 2.2 x 10^-4 How is this done?
2017-04-25 1:02 pm
回答 (2)
2017-04-25 1:31 pm
✔ 最佳答案
HCNO(aq) + H₂O(l) ⇌ CNO⁻(aq) + H₃O⁺(aq) .... Ka = 2.2 × 10⁻⁴pH = pKa + log([CNO⁻]/[HCNO]
pH = -log(2.2 × 10⁻⁴) + log(0.10/0.25)
pH = 3.26
The answer: d.) 3.26
2017-04-25 1:35 pm
HOCN <======> H+ + OCN-
Make an ICE table, at the equilibrium , the concentration of [ HOCN] = ( 0.25 - x) M ; [ H+] = x; [OCN- ] = 0.10 + x => ka = (x)( 0.10 + x )/( 0.25 - x-Since x small => Ka = (x)( 0.10 ) / ( 0.25 ) =>
2.2 * 10^-4 = 0.1 x/0.25 => x = 5.5 * 10^-4
pH = - log [H+] = - log ( 5.5 * 10^-4 ) = 3.26
Make an ICE table, at the equilibrium , the concentration of [ HOCN] = ( 0.25 - x) M ; [ H+] = x; [OCN- ] = 0.10 + x => ka = (x)( 0.10 + x )/( 0.25 - x-Since x small => Ka = (x)( 0.10 ) / ( 0.25 ) =>
2.2 * 10^-4 = 0.1 x/0.25 => x = 5.5 * 10^-4
pH = - log [H+] = - log ( 5.5 * 10^-4 ) = 3.26
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