question 9a-b 2 9, a. A 0.411 g sample of KHP is dissolved with 50.00 ml. of dei
ID: 550125 • Letter: Q
Question
question 9a-b
2 9, a. A 0.411 g sample of KHP is dissolved with 50.00 ml. of deionized H,O in a 125-ml Erlenmeyer flask. The sample is titrated to the endpoint with 15.17 ml. of a NaOH solution. What is the molar concentration of the NaOH solution? Give answer to the correct number of significant figures. b. A 25.00 mL aliquot of nitric acid (HNOs) of unknown concentration is pipetted into a 125- mL Erlenmeyer flask and 2 drops of phenolphthalein are added. The above NaOH (titrant) solution is used to titrate the HNOs (analyte). If 16.77 mL of titrant is used to reach the endpoint, what is the molar concentration of the nitric acid? Give answer to the correct number of significant figures.Explanation / Answer
9a. KHP will react only with 1 mole of NaOH
1 KHP + 1 NaOH ==== 1 KNaP + 1 H2O
First let´s get molecular weight of KHP, from internet mw = 204.2 g/gmol
moles = mass / molecular weight = 0.411 / 204.2 = 0.00201 moles of KHP available
So 0.00201 moles of KHP according to stoichiometry can only react with 0.00201 moles NaOH
Molarity = moles / Volume
Molarity = 0.00201 / 0.01517 L = 0.133 M
9b
We will use the same molarity from our first example
Molarity of NaOH = 0.133 M
Volume required = 16.77 ml = 0.01677 L
moles = Molarity * Volume = 0.133 * 0.01677 = 0.0022 moles of NaOH available
1 mole of HNO3 can only react with 1 mole of NaOH so
if we made react 0.0022 moles of NaOH then we know we have 0.0022 moles of HNO3
Molarity of nitric acid is (with the volume of 25 ml = 0.025 L)
0.0022 moles / 0.025 L = 0.088 M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.