A Student Performing a lab weighed a 1.0993g sample, containing an unknown KHP, which required 18.06mL of 0.1879M NAOH to reach the phenolphtalein endpoint.

a) How many moles of base were consumed by the KHP in this sample?
b) How many grams of KHP were titrated?
c)What is the mass percent of KHP in the sample?

mols NaOH = M x L = ?

mols KHP = same (1:1 ratio) from
KHP + NaOH ==> KNaP + H2O.

Calculate mass KHP from g = mols x molar mass.

Then %KHP = (mass KHP/mass sample)*100 = ?

To find the answers to these questions, we need to follow a step-by-step process using the given information and some basic calculations. Let's break it down:

a) How many moles of base were consumed by the KHP in this sample?

Step 1: Calculate the number of moles of base used:
Moles of base = concentration of NaOH x volume of NaOH used (in liters)
Moles of base = 0.1879 M x 0.01806 L
Moles of base = 0.003394874 moles

b) How many grams of KHP were titrated?

Step 2: Identify the mole ratio between KHP and the base used:
From the balanced equation of the reaction between KHP and NaOH:
KHP + NaOH -> KNa + H2O + CO2
We can see that the mole ratio between KHP and NaOH is 1:1.

Step 3: Use the mole ratio to find the moles of KHP:
Moles of KHP = 0.003394874 moles (since it's a 1:1 ratio)

Step 4: Find the molar mass of KHP:
Molar mass of KHP = (molar mass of K) + (molar mass of H) + (molar mass of P) + (4 * molar mass of O)
Molar mass of KHP = 39.1 g/mol + 1.01 g/mol + 30.97 g/mol + (4 * 16.00 g/mol)
Molar mass of KHP = 204.23 g/mol

Step 5: Calculate the grams of KHP:
Grams of KHP = Moles of KHP x Molar mass of KHP
Grams of KHP = 0.003394874 moles x 204.23 g/mol

c) What is the mass percent of KHP in the sample?

Step 6: Determine the mass of the sample:
Mass of the sample = 1.0993 g (given in the question)

Step 7: Calculate the mass percent of KHP:
Mass percent of KHP = (grams of KHP / Mass of the sample) x 100
Mass percent of KHP = (0.003394874 moles x 204.23 g/mol) / 1.0993 g x 100

By following these steps, you can find the answers to each question.