When a 5.00 ml sample of vinegar is titrated, 44.5 ml of 0.100 m NAOH solution is required to reach the end point. What i sthe concentration of vinegar in moles per liter of solution?
To find the concentration of vinegar in moles per liter of solution, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between vinegar (acetic acid) and sodium hydroxide (NaOH).
The balanced equation for the neutralization reaction is:
CH3COOH (acetic acid) + NaOH (sodium hydroxide) → CH3COONa (sodium acetate) + H2O
From the balanced equation, we can see that the molar ratio between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is 1:1. This means that for every 1 mole of acetic acid reacted, 1 mole of sodium hydroxide is required.
Now, let's calculate the number of moles of sodium hydroxide used in the titration. The volume of the sodium hydroxide solution used is 44.5 ml, which is equivalent to 0.0445 L (since 1 ml = 0.001 L). The concentration of the sodium hydroxide solution is 0.100 M, which means it contains 0.100 moles of NaOH per liter of solution.
Using the formula:
moles = concentration × volume
moles of NaOH = 0.100 M × 0.0445 L = 0.00445 moles
Since the molar ratio between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is 1:1, the number of moles of acetic acid in the vinegar sample is also 0.00445 moles.
To find the concentration of acetic acid in moles per liter of solution, we need to consider the volume of the vinegar sample. The sample size is 5.00 ml, which is equivalent to 0.00500 L.
Now, we can calculate the concentration of acetic acid (vinegar) in moles per liter:
concentration = moles/volume
concentration = 0.00445 moles / 0.00500 L = 0.890 M
Therefore, the concentration of vinegar (acetic acid) in moles per liter of solution is 0.890 M.
HAc + NaOH ==> NaAc + H2O
mols NaOH = M x L = ?
mols HAc = mols NaOH (look a the coefficients in the equation).
Then mol/L soln = M.