According to HomeworkLib rules we have to answer only first one problem
HCOONa >> HCOO- + Na+
HCOO- + H2O <----> HCOOH + OH-
K = Kw/Ka = 1 x 10^-14 / 1.8 x 10^-4 = x^2/ 0.25-x
x = [OH-]= 3.73 x 10^-6 M
pOH = 5.43
pH =14- p OH = 8.57
OR
pH=7+0.5 [pKa+log(C) ]
where C is the concentration of the salt.
so pH=7+ 0.5*3.72 +0.5 log(0.25)
= 7+ 1.86 -0.3010
=8.559
Percentage ionization = [OH-] / B *100
= 3.73 x 10^-6/0.25 *100
= 0.001492%
=0.0015 %
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