1b. Calculate the exact molarity of a solution of NaOH if 55.00 mL of it is need

Question

1b. Calculate the exact molarity of a solution of NaOH if 55.00 mL of it is needed to titrate an amount of KHP that equals your average mass (that you stated above). Average is 50.0 (4 pts)

2a. What is the average molarity for your standardized NaOH solution?_______________ (1 pt) Average: .0075 M

2b. If 1.20 grams of impure solid KHP sample required 2.53 mL of your standardized NaOH to reach the end point, what was the percent KHP in this sample? (4 pts)

3. A buffer solution is prepared by adding 1.5 g of potassium acetate (FW=98.15 g/mol) to 5.80 mL of 3.00 M acetic acid. Water was added to bring the total volume to 100.0 mL.   The pH of the buffer was then adjusted to a value of pH = 4.50 using 2.00 M HCl. How many mL of HCl were added in order to make this adjustment? (6 pts)

Explanation / Answer

1. You want to titrate a mass of 50 grams of KHP, first you need to get the molar mass of KHP, the mw is 204.22 g/gmol

moles of KHP = mass of KHP / molar mass = 50 / 204.22 = 0.24483 moles of KHP available

we also know that the reaction between KHP and NaOH is 1:1

KHC8H4O4 + NaOH ----> KNaC8H4O4 + H2O

so 0.24483 moles of KHP will necessarily need the same ammount of NaOH so we need

0.24483 moles of NaOH

to calculate molarity you use Molarity = moles / volume, use the 55 ml from the statement, put them in liters (0.055L):

Molarity = 0.24483 / 0.055 L = 4.45 M

*Just 1 question at a time please =)

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