A 0.144 M solution of a monoprotic acid has a percent dissociation of 1.60%. what is Ka?

Question

A 0.144 M solution of a monoprotic acid has a percent dissociation of 1.60%. What is the acid dissociation constant (Ka) for this acid?

Anonymous · Accepted Answer

HA —-> H+ + A-
Ka = [H+]*[A-] / [HA]
[H+] = [A-] = Ca*α
[HA] = Ca*(1 – α)
α = dissociation % / 100 = 0.0160
so
Ka = (0.144*0.0160)^2 / (0.144*(1 – 0.0160)) = 3.75*10^-5

Anonymous · Answer

assume a weak monoprotic acid HA. HA ? H(+) + A(-) at t=0 C 0 0 at eq. C(one million-x) Cx Cx Now, [HA] initially is C(=0.174 M). If ‘x’ is the degree of dissociation of acid, at equilibrium, [HA] very last will be C-Cx, i.e. C(one million-x), and [H(+)] & [A(-)] formed could be Cx. Now, Ka = [A(-)][H(+)]/[HA] => Ka = Cx²/(one million-x) Now, on condition that Ka = 2.5×10^-3 and for once you concentrate on that answer is dilute, we would properly be waiting to approximate (one million-x) as one million. therefore, inducing the above reported consequences and value of C as 0.174M, we get 2.5×10^-3 = 0.174x² => x ? 0.11987 therefore, p.c.. ionisation = 11.987% ? 12% you should as suitable analyze the effect by no longer utilizing the approximation. if so a quadratic equation will be formed. you receives 2 roots, one useful and one unfavorable. ignore the unfavorable one. by that technique, you get x ? 0.1129

Anonymous · Answer

ka is pottasium

Anonymous

A 0.144 M solution of a monoprotic acid has a percent dissociation of 1.60%. what is Ka?

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