If 34.7 mL of 0.210 M KOH is required to completely neutralize 30.0 mL of a CH3COOH solution, what is the molarity of the acetic acid solution?
When the molecules react 1:1 as they do in this case you can use
mL acid x M acid = mL base x M base.
what volume of 1.90 M SrCl2 is needed to prepare 525mL of 5.00mM SrCl2?
To find the molarity of the acetic acid solution, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between KOH and CH3COOH.
The balanced chemical equation for the reaction is:
CH3COOH + KOH -> CH3COOK + H2O
From the equation, we can see that the mole ratio between KOH and CH3COOH is 1:1. This means that for every 1 mole of KOH, there is 1 mole of CH3COOH.
We are given that 34.7 mL of 0.210 M KOH is required to neutralize 30.0 mL of the acetic acid solution.
First, let's calculate the number of moles of KOH used:
moles of KOH = volume of KOH (in L) * molarity of KOH
moles of KOH = 0.0347 L * 0.210 mol/L = 0.007287 mol KOH
Since the mole ratio between KOH and CH3COOH is 1:1, the number of moles of CH3COOH in the acetic acid solution is also 0.007287 mol.
Now, let's calculate the molarity of the acetic acid solution:
molarity of CH3COOH = moles of CH3COOH / volume of CH3COOH (in L)
molarity of CH3COOH = 0.007287 mol / 0.0300 L = 0.2429 M
Therefore, the molarity of the acetic acid solution is 0.2429 M.