A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentrations...
A monoprotic weak acid, HA, dissociates in water according to the reaction
HA(aq)↽−−⇀H+(aq)+A−(aq)
The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=2.00×10−4 M, and [A−]=2.00×10−4 M. Calculate the value of pKa for the acid HA.
Homework Answers
Ka = [H+][A-]/[HA]
= (2.00*10^-4)*(2.00*10^-4)/(0.280)
= 1.43*10^-7
use:
pKa = -log Ka
= -log (1.43*10^-7)
= 6.84
Answer: 6.84
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A monoprotic weak acid, HA, dissociates in water according to
the reaction
HA(aq)↽−−⇀H+(aq)+A−(aq)
The equilibrium concentrations...
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A monoprotic acid, HA, is dissolved in water: HA <------> H+ +A- The equilibrium concentrations of the reactants and products are [HA] = 0.280 M [H ] = 2.00 × 10–4 M [A–] = 2.00 × 10–4 M. Calculate the value of pKa for the acid HA.
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help A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) = (aq)...
help
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A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) – H(aq) + A (aq) The equilibrium concentrations of the reactants and products are [HA] = 0.270 M, [H+] = 4.00 x 10-4 M, and [A-] = 4.00 x 10-4 M. Calculate the value of pKa for the acid HA. pKa =
A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) + H2O(1) H2O+(aq) + A-(aq) The equilibrium concentrations of the reactants and products are [HA] = 0.250 M, H,O+] = 2.00 x 10-4 M, and [A-] = 2.00 x 10-4 M. Calculate the Ka value for the acid HA. Ka = 6.79588
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