A sample of sodium hydroxide of volume 11 mL was titrated to the stoichiometric point with 24.5 mL of 0.138 M HCl(aq). What is the initial molarity of NaOH in the solution?
Answer in units of M.
.011*m=.0245*.138
solve for molarity
To determine the initial molarity of NaOH, we can use the equation:
M1V1 = M2V2
Where:
M1 is the initial molarity of NaOH
V1 is the initial volume of NaOH solution
M2 is the molarity of HCl(aq)
V2 is the volume of HCl(aq) required to reach the stoichiometric point
Plugging in the given values:
M1 * 11 mL = 0.138 M * 24.5 mL
Rearranging the equation:
M1 = (0.138 M * 24.5 mL) / 11 mL
Calculating the value:
M1 = 0.306 M
Therefore, the initial molarity of NaOH in the solution is 0.306 M.
To find the initial molarity of NaOH in the solution, we need to use the concept of stoichiometry and the given volumes and concentrations of the reactants in the titration.
First, let's write down the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
From the equation, we can see that the stoichiometric ratio between NaOH and HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.
Next, let's determine the number of moles of HCl that reacted in the titration. We can use the formula: moles = concentration x volume.
Moles of HCl = 0.138 M x 24.5 mL
= 0.138 mol/L x 24.5 x 10^(-3) L
= 0.003381 mol
Since the stoichiometric ratio between NaOH and HCl is 1:1, the number of moles of NaOH in the solution is also 0.003381 mol.
Finally, let's calculate the initial molarity of NaOH. Molarity is defined as moles of solute divided by the volume of the solution in liters.
Initial molarity of NaOH = moles of NaOH / volume of NaOH solution in liters
= 0.003381 mol / 11 mL x 10^(-3) L/mL
= 0.307 M
Therefore, the initial molarity of NaOH in the solution is 0.307 M.