A sample of 1.018 g of KHP (potassium hydrogen phthalate, molar mass = 204.22 g/mol) was dissolved in ~ 25 mL distilled water and titrated with a NaOH solution of unknown concentration.
If 28.69 mL of base was used to reach the endpoint, what was the concentration of the NaOH (in mol/L)?
mool KHP = grams/molar mass
mols NaOH = mols KHP (from the coefficients in the balanced equation.)
M NaOH = mols NaOH/L NaOH
So does [mols NaOH = mols KHP] mean that the values of one is equal to the other?
yes.
Write the equation and look at the coefficients.
KHP + NaOH ==> NaKP + H2O
1 mol KHP = 1 mol NaOH
To find the concentration of the NaOH solution, we need to use the concept of stoichiometry and the equation of the reaction between KHP and NaOH.
The balanced equation for the reaction between KHP and NaOH is as follows:
KHP + NaOH → KNaP + H2O
From the equation, we can see that the stoichiometric ratio between KHP and NaOH is 1:1. This means that 1 mole of KHP reacts with 1 mole of NaOH.
First, let's find the number of moles of KHP using its mass and molar mass:
Given:
Mass of KHP = 1.018 g
Molar mass of KHP = 204.22 g/mol
Number of moles of KHP = Mass of KHP / Molar mass of KHP
= 1.018 g / 204.22 g/mol
≈ 0.00499 mol
Since the stoichiometry between KHP and NaOH is 1:1, the number of moles of NaOH used in the titration is also 0.00499 mol.
Next, we need to find the concentration of NaOH in mol/L (Molarity).
Given:
Volume of NaOH used = 28.69 mL = 0.02869 L
Concentration of NaOH (in mol/L) = Number of moles of NaOH / Volume of NaOH used
= 0.00499 mol / 0.02869 L
≈ 0.174 M
Therefore, the concentration of the NaOH solution is approximately 0.174 mol/L.