A 25,0ml sample of 0,105M HCl was tritrated with 31,5ml of NaOH.what is the concentration of the NaOH?
millimols HCl = mL x M = ?
millimols NaOH used = mmols HCl since 1 mol HCl = 1 mol NaOH in the equation.
Then M NaOH = mmols/mL = ?
The equation is HCl + NaOH ==> NaCl + H2O
Post your work if you get stuck.
To find the concentration of NaOH, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and NaOH. The balanced equation is:
HCl + NaOH -> NaCl + H2O
The stoichiometry of the reaction tells us that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.
Given:
Volume of HCl (VHCl) = 25.0 mL = 0.0250 L
Molarity of HCl (MHCl) = 0.105 M
Volume of NaOH (VNaOH) = 31.5 mL = 0.0315 L
To calculate the concentration of NaOH, we can use the following equation, which comes from the stoichiometry of the reaction:
MHCl * VHCl = MNaOH * VNaOH
Rearranging the equation to solve for MNaOH:
MNaOH = (MHCl * VHCl) / VNaOH
Plugging in the values:
MNaOH = (0.105 M * 0.0250 L) / 0.0315 L
MNaOH ≈ 0.0833 M
Therefore, the concentration of NaOH is approximately 0.0833 M.