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 = ??
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.
4 answers
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.
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!!
2 answers