a solution of sulfuric acid was titrated with sodium hydroxide. if you took 23.7 mL of .025 M NaOH to reach the endpoint with a 5.0 mL sample of the acid what was its concentration?
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols NaOH = M x L = ?
mols H2SO4 = 1/2 mols NaOH (look at the coefficients in the balanced equation).
M H2SO4 = mols H2SO4/L H2SO4
To calculate the concentration of the sulfuric acid, we can use the concept of stoichiometry and the balanced chemical equation between sulfuric acid (H2SO4) and sodium hydroxide (NaOH).
The balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the balanced equation, we can see that one mole of sulfuric acid (H2SO4) reacts with two moles of sodium hydroxide (NaOH).
Given:
Volume of NaOH used (V1) = 23.7 mL
Molarity of NaOH (M1) = 0.025 M
Volume of acid sample (V2) = 5.0 mL
Unknown concentration of H2SO4 (M2) = ?
To find the concentration (M2) of sulfuric acid, we can use the following formula:
M1V1 = M2V2
Substituting the given values:
(0.025 M) * (23.7 mL) = M2 * (5.0 mL)
Now, let's solve for M2:
M2 = (0.025 M * 23.7 mL) / 5.0 mL
M2 = 0.1185 M
Therefore, the concentration of the sulfuric acid solution is 0.1185 M.