NaOH and HCl react 1:1 so you can use the simplified equation
mL NaOH + M NaOH = mL HCl x M HCl
mL NaOH + M NaOH = mL HCl x M HCl
The balanced chemical equation for the reaction is:
HCl + NaOH -> NaCl + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of water (H2O).
Given:
Volume of HCl solution (V1) = 50.0 mL = 0.050 L
Concentration of NaOH (C2) = 0.650 M
Volume of NaOH solution (V2) = 90.0 mL = 0.090 L
To find the moles of NaOH used in the reaction, we can use the formula:
moles of solute = concentration × volume
moles of NaOH (n2) = C2 × V2
= 0.650 M × 0.090 L
Now, since the stoichiometry of the reaction is 1:1 between NaOH and HCl, the moles of HCl (n1) are equal to the moles of NaOH used.
n1 = n2
= 0.650 M × 0.090 L
Since we know the volume of HCl solution (V1), we can calculate the molarity (M1) using the formula:
Molarity = moles of solute / volume of solution
M1 = n1 / V1
= (0.650 M × 0.090 L) / 0.050 L
Now we can plug in the values and calculate:
M1 = (0.0585 mol) / (0.050 L)
M1 ≈ 1.17 M
Therefore, the molarity of the HCl solution is approximately 1.17 M.