Question

An unknown sample (0.6855 g) containing khp is titrated to the equivalence point with 9.75 ml of 0.07892 M NaOH. a. How many moles of khp are present? b. How many grams of khp are present (molecular weight of khp = 204.23 g/mole)? c. What is the percent khp in the sample?

Answer 1

The number of moles of KHP present is 0.76827 moles. The grams of KHP present is 156.92 grams. The percent KHP in the sample is 228.81%.

Step-by-step explanation:

To find the number of moles of KHP present, we can use the equation:

moles of KHP = (volume of NaOH) x (molarity of NaOH)

Plugging in the values, we get:

moles of KHP = (9.75 mL) x (0.07892 mol/L)

moles of KHP = 0.76827 moles

To find the grams of KHP present, we can use the equation:

grams of KHP = moles of KHP x (molar mass of KHP)

Plugging in the values, we get:

grams of KHP = 0.76827 moles x 204.23 g/mol

grams of KHP = 156.92 grams

To find the percent KHP in the sample, we can use the equation:

percent KHP = (mass of KHP / mass of sample) x 100%

Plugging in the values, we get:

percent KHP = (156.92 grams / 0.6855 grams) x 100%

percent KHP = 228.81%