I'm having trouble with these questions. Could someone help me?
1) If 20.6 mL of 0.010 M aqueous HCl is required to titrate 30.0 mL of an aqueous solution of NaOH to the equivalence point, what is the molarity of the NaOH solution?
2) In the titration of 35.0 mL of drain cleaner that contains NaOH, 50.08 mL of 0.409 M HCl must be added to reach the equivalence point. What is the concentration of the base in the cleaner?
3) Titrating a sludge sample of unknown origin required 41.55 mL of 0.1125 M NaOH. How many moles of H3O+ did the sample contain?
4) Neutralizing 5.00L of an acid rain sample required 11.3 mL of 0.0102 M KOH. Calculate the hydronium ion concentration in the rain sample.
moles HCl = 0.0206 L x 0.010 M = 0.000206
at the equivalence point : moles acid = moles base
moles NaOH = 0.000206
Molarity NaOH = 0.000206 / 0.0300 L= 0.00687 M
moles HCl = 0.05008 L x 0.409 M=0.0205
molarity NaOH = 0.0205/ 0.0350 L=0.585 M
moles NaOH = 0.04155 L x 0.1125 M=0.004674
moles H3O+ = 0.004674
moles KOH = 0.0113 L x 0.0102 M =0.000115
[H+]= 0.000115 / 5.00 L=2.31 x 10^-5 M
In each case, since you are going to the equivalence point, the mol of acid and base are equal.
#1) 20.6 mL X 0.01 mol/L = 30.0 mL X (unknown concentration)
Same deal for the others.