if 50.0 mLs of HCl of unknown concentration is titrated to neutrality with 90.0 mL of .650 M NaOH. calculate the molarity of the HCl

NaOH and HCl react 1:1 so you can use the simplified equation

mL NaOH + M NaOH = mL HCl x M HCl

1.17M

To calculate the molarity of HCl, we can use the concept of stoichiometry and the balanced chemical equation of the neutralization reaction between HCl and NaOH.

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.