Asked by Nina
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
Answered by
Bot
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).
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