Question
A student completes titration by adding 12.0 mL of NaOH(aq) of an unknown concentration to 16.0 mL of 0.15M HCl(aq). what is the molar concentration of the NaOH(aq)?
Answers
The easy way to do this is
mL NaOH x M NaOH = mL HCl x M HCl
This equation works only when the reactants are 1:1 in the equation as is the case with HCl and NaOH.
NaOH + HCl ==> NaCl + H2O
mL NaOH x M NaOH = mL HCl x M HCl
This equation works only when the reactants are 1:1 in the equation as is the case with HCl and NaOH.
NaOH + HCl ==> NaCl + H2O
0.11M
Using the formula:
mL NaOH x M NaOH = mL HCl x M HCl
We can plug in the values and solve for M NaOH:
12.0 mL x M NaOH = 16.0 mL x 0.15 M
M NaOH = (16.0 mL x 0.15 M) / 12.0 mL
M NaOH = 0.20 M / 12.0 mL
M NaOH = 0.0167 M
Therefore, the molar concentration of the NaOH(aq) is 0.0167 M (or approximately 0.02 M).
mL NaOH x M NaOH = mL HCl x M HCl
We can plug in the values and solve for M NaOH:
12.0 mL x M NaOH = 16.0 mL x 0.15 M
M NaOH = (16.0 mL x 0.15 M) / 12.0 mL
M NaOH = 0.20 M / 12.0 mL
M NaOH = 0.0167 M
Therefore, the molar concentration of the NaOH(aq) is 0.0167 M (or approximately 0.02 M).