Question 26A student completes a titration by adding 12.0 milliliters of NaOH(aq) of unknown concentration to 16.0 milliliters of 0.15 M HCl(aq). What is the molar concentration of the NaOH(aq)?
0.2
mL x M = mL x M
To find the molar concentration of NaOH(aq), we can use the balanced equation for the reaction:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
According to the balanced equation, the ratio of NaOH to HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.
Given that the volume of NaOH(aq) used is 12.0 milliliters and the concentration of HCl(aq) is 0.15 M, we can use the equation:
Moles of NaOH = Volume of NaOH(aq) × Concentration of HCl(aq)
Moles of NaOH = 12.0 mL × 0.15 M = 1.8 millimoles
Since the ratio of NaOH to HCl is 1:1, the number of moles of NaOH is equal to the number of moles of HCl. Therefore, the molar concentration of NaOH(aq) is 1.8 millimoles divided by the volume of NaOH used in liters.
To convert milliliters to liters, we divide by 1000:
Volume of NaOH in liters = 12.0 mL / 1000 = 0.012 L
Molar concentration of NaOH(aq) = Moles of NaOH / Volume of NaOH in liters
Molar concentration of NaOH(aq) = 1.8 millimoles / 0.012 L = 150 millimoles per liter, or 0.150 M
Therefore, the molar concentration of NaOH(aq) is 0.150 M.