Here is how you do these; I'll do the first one.
9.00 mL x 0.100M HCl = 0.900 mmols.
100.0 mL x 0.100 M NH3 = 10.0 mmols.
100.0 mL x 0.100 M NH4Cl = 10.0 mmols.
...........NH3 + H^+ ==> NH4^+
I.........10.....0.......10
add.............0.9..............
C.......-0.9...-0.9......+0.9
E.........9.1...0.........10.9
pH = pKa + log(base)/(acid)
What are you using for pKa? Use the one in your text and not what I've used.
pH = 9.24 + log(9.1/10.9) = 9.16.
The question asks for CHANGE in pH. It was 9.24 before the HCl was added. Change is 9.24 to 9.16 = -0.08. Check my work. Watch the significant figures; I didn't.
calculate the change in pH when 9.00ml of 0.100M HCl(aq) is added to 100.0mL of a buffer solution that is 0.100M in NH3(aq) and 0.100M NH4Cl(aq).
...Calculate the change in pH when 9.00,L if 0.100M NaOH(aq) is added to the original buffer solution.
1 answer
1 answer