Once I have the molarity for KHP which is 0.0027 M, and i add 14.48 mL of NaOH to the solution, what is the concentration of NaOH in the solution.
I wonder if you REALLY have the molarity of KHP. Usually that is a solid and one weighs it as a solid and titrates the entire sample; thus, moles is the number we needed. However, if you mean what you say, then mL x M = mL x M.
Since you don't list a volume for KHP, I suspect 0.0027 is moles, then, since NaOH and KHP react 1:1, we know moles NaOH = 0.0027, also, and
M NaOH = 0.0027/0.1448 = ??
Did i get it wrong? It has KHP molar mass at 204.33 g/mol and we use 0.556 g of KHP dissolved in water. THen if 14.48 mL of NaOH is used to reach endpoint, it needs the concentration of NaOH?
No, you did it right but you labeled it wrong.
moles KHP = 0.556 g KHP/204.33 = 0./00272 ( I wouldn't throw away the last 2).
Then since 1 mole KHP = 1 mole NaOH, then
M NaOH = moles NaOH/L NaOH = 0.00272/0.01448 = 0.1879 which to three significant figures rounds to 0.188 M.
Your only error was labeling 0.0027 as M KHP and not as moles KHP.
thank you for your help!!
To find the concentration of NaOH in the solution, you can use the concept of stoichiometry and the equation of the balanced chemical reaction between KHP (potassium hydrogen phthalate) and NaOH.
First, let's write the balanced chemical equation for the reaction:
KHP + NaOH -> NaKP + H2O
From the balanced equation, we can see that the ratio of KHP to NaOH is 1:1.
Given that the molarity of KHP is 0.0027 M, we can determine the number of moles of KHP present in the solution using the formula:
Moles of KHP = molarity of KHP x volume of KHP (in liters)
Volume of KHP = (14.48 mL) / 1000 = 0.01448 L
Moles of KHP = 0.0027 M x 0.01448 L = 0.000039216 moles
Since the ratio of KHP to NaOH is 1:1, the number of moles of NaOH present in the solution is also 0.000039216 moles.
Now, let's calculate the concentration of NaOH in the solution:
Concentration of NaOH = moles of NaOH / volume of NaOH (in liters)
Volume of NaOH = 14.48 mL / 1000 = 0.01448 L
Concentration of NaOH = 0.000039216 moles / 0.01448 L ≈ 0.00271 M
Therefore, the concentration of NaOH in the solution is approximately 0.00271 M.