For those acids that dissociate completely, mol HNO3/L = (H^+) and convert that to pH.
For Ca(OH)2, same thing but (OH^-) = 2 x M Ca(OH)2, convert OH^- to pOH then to pH.
For weak acids with a pKa, convert pKa to Ka. Then let me abbreviate chloroacetic acid as HA.
..........HA ==> H^+ + A^-
I......0.180M....0.....0
C........-x......x......x
E......0.180-x...x......x
Ka = (H^+)(A^-)/(HA) Substitute and solve for x = H^+) and convert to pH.
For the sodium chloracetate, 0.020M, this is the hydrolysis of the salt; therefore,
.........A^- + HOH ==> HA^- + OH^-
I.......0.020..........0.......0
C........-x............x.......x
E......0.020-x.........x.......x
Kb for acetate ion = (Kw/Ka for chloroacetic acid) = (x)(x)/(0.020-x) and solve for x = (OH^-), convert that to pH.
Post your work if you get stuck.
the problem is find the pH of a solution prepared by dissolving all of the following in sufficient water to yield 1.00 L of solution :0.180 mole of chloroacetic acid (CICH2CO2H),0.020 mole ofCICH2CO2Na,0.080 mole HNO3, and 0.080 mole Ca(OH)2. Assume that the Ca(OH)2 and HNO3 dissociate completely and that the pka of chloroacetic acid is 2.86.
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