a and b are salts. The pH of those solutions is determined by the hydrolysis of the salt. c is a mixture; you calculate the molarity of the acetate ion and go from there. Here is how you do the first one, in detail.
.............F + HOH ==> HF + OH^-
initial.....0.1..........0......0
change.......-x..........x.......x
equil.......0.1-x.........x......x
Kb for F^- = (Kw/Ka for HF) = (HF)(OH^-)/(HF) = (x)(x)/(0.1-x)
and solve for x = (OH^-) = (HF) and convert to pH.
A shortcut I use for calculating pH and pOH and that stuff is, if x is OH^-, as it will be in all of these calculations, I take pOH = -log(OH^-), then convert to pH by
pH + pOH = pKw = 14. So while the pOH is still in the calculator (as a negative number), and without changing the sign, I key in + 14 and out pops the pH.
1)calculate [OH-] and pH for each of the following solutions. (Kw = 1.0*10^-14.)
a)0.101 M NaF
b) 0.030 M Na2S
c)a mixture that is 0.019 M in CH3COONa and 0.036 M in (CH3COO)2Ba
please help and if u can try to explain
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