Calculate soln. pH if 100mL 0.10M Na2S is mixed with 200mL 0.050M HCL. (HINT: Consider Kb for HS-[1.1x10-7])?
Please help explain how to arrive at this answer! Any help would be great! Thanks!
Answer:
Na2S is a salt of NaOH and H2S.
So when you add HCl, it will go on to react with NaOH.
Moles of NaOH = 0.02mol
Moles of HCl = 0.01mol
So almost all the Na2S is converted to NaHS. This is the salt of weak cation Na(+) and strong anion HS(-).
So we can use the formula
pH=7+1/2(pKb+log c){c is the concentration = 0.01/0.3 = 0.033}
pH=7+0.5(6.958 + log 0.033)
pH=7+0.5(6.958 - 1.481)
pH=7+2.73=9.73
The further deficit of 0.01 can be accounted for by considering the dissociation of HS(-) to S(-2),which is almost negligible. For that you will need Kb of S(-2)
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