What is the pH of the solution created by combining 1.90 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
mL NaOH pH with HCl pH with HC2H3O2
1.90 ? ?
There are two problems here.
#1. pH when adding 1.90 mL of 0.1 M NaOH to 8.00 mL of 0.1 M HCl.
moles NaOH = M x L = 0.0019 x 0.1 = 0.00019 moles
moles HCl = 8.00 mL x 0.1 M = 0.008 x 0.1 = 0.0008.
NaOH + HCl ==> NaCl + H2O
Place the moles below the reactants so you can see what is going on.
0.00019 moles NaOH + 0.0008 moles HCl =
NaOH used up. We form 0.00019 moles NaCl, 0.00019 moles H2O, and 0.0008-0.00019=0.00061 moles HCl left over and pH = -log(H^+) = etc.
#2. 1.90 mL of 0.1M NaOH + 8.00 mL of 0.1 M HC2H3O2.
Write the equation as above. Calculate moles of each from M x L. Plug in the NaC2H3O2 formed, H2O formed, and the excess reagent remaining. You should get these numbers
NaC2H3O2 formed = 0.00019 moles
H2O formed = 0.00019 moles.
NaOH all used.
HC2H3O2 = 0.0008-0.00019 = 0.00061 moles HC2H3O2 remaining.
So you have a weak acid (HC2H3O2) and its salt (NaC2H3O2) which is a buffered solution. Use the Henderson-Hasselbalch equation to solve for pH. If I didn't goof the answer should be pH = 4.23.