Question

Greetings,

working on a buffer lab report,

pH of 50.0 mL H2O + 1.00 mL of 0.6 M NaOH = ?

PH of 50.0mL H2O + 1.00 mL of 0.6 HCl = ?

don't think from the lack of informatino given you can help me with these two, but the others are

the pH of acetate buffer

and pH of ammonia buffer

please show step by step instructions, any answers are appreciated.

Answer 1

pH of 50.0 mL H2O + 1.00 mL of 0.6 M NaOH

mol of NaOH = MV = 1*0.6 = 0.6 mmol

[OH-] = mmol/total mL = (0.6)/(50+1) =0.011764 M

pOH = -log(OH) = -log(0.011764) = 1.92944

pH = 14-pOH = 14-1.92944 = 12.07056

PH of 50.0mL H2O + 1.00 mL of 0.6 HCl =

similar...

mol of HCl= MV = 1*0.6 = 0.6 mmol

[H+] = mmol/total mL = (0.6)/(50+1) =0.011764 M

pH = -log(H) = -log(0.011764) = 1.92944

for a pH of acetate buffer you need:

acetate formation, from the weak acid

for the ammonia buffer you need

ammonia ion NH4+ from the NH3 in solution