posted by canas .
Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added and make a graph of pH versus Va: Va = 0,6 and 12 mL
HBr + NaOH ==> NaBr + H2O
1. First you need to calculate the volume of HBr required to reach the equivalence
mols NaOH = M x L = 0.1M x 0.1L =- 0.01
mols HBr must be 0.01 to reach the equivalence point.
Then M HBr = mols HBr/L HBr. Substitute to obtain 1M = 0.01/L and L = 0.01/1 = 0.01L or 10.0 mL of the 1.00 M HBr. That tells you that the 6.0 mL HBr is BEFORE the eq. pt. and 12.0 mL HBr is AFTER the eq. pt.
a. at zero HBr.
You have 0.1M NaOH; therefore, (OH^-) = 0.1. Convert that to pH.
b. mols NaOH initially = 0.1M x 0.1L = 0.01.
mols HBr added at 6 mL = 1M x 0.006 L = 0.006 mols
0.01 mol NaOH - 0.006 mol HBr = 0.004 and since M = mols/L then 0.004/(0.106) = 0.0377. That is OH^-, convert to pH.
I will leave c for you. That is after the equivalence point.