Question

A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is instructed to pipet a 20.0 mL portion of the strong base solution into a conical flask, to add two drops of an indicator that changes at pH = 7, and to dispense the standard H2SO4 solution from a buret until the solution undergoes a permanent color change. The initial buret reading is 5.00 mL and the final reading is 30.00 mL at the equivalence point.Which procedural error will result in a strong base molarity that is too high?Using 4 drops of indicator rather than the recommended 2 dropsUsing a buret with a tip filled with air rather than the H2SO4 solutionUsing an indicator that changes at pH = 5 rather than at pH = 7Using a conical flask that contains several drops of H2O

Answer 1

The molarity of the strong base is 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

Step-by-step explanation:

Step 1: Data given

Molarity of H2SO4 = 0.250 M

The initial buret reading is 5.00 mL

The final reading is 30.00 mL

Step 2: Calculate volume of H2SO4 used

30.00 mL - 5.00 mL = 25.00 mL

Step 3: Calculate moles of H2SO4

0.250 M = 0.250 mol/L

Since there are 2 H+ ions per H2SO4

0.250 mol/L * 2 = 0.500 mol/L

The number of moles H2SO4 = 0.500 mol/L * 0.025 L

Number of moles H2SO4 = 0.0125 mol

Step 4: Calculate moles of OH-

For 1 mol H2SO4, we need 1 mol of OH-

For 0.0125 mol of H2SO4, we have 0.0125 mol of OH-

Step 5: Calculate the molarity of the strong base

Molarity = moles / volume

Molarity OH- = 0.0125 mol / 0.02 L

Molarity OH - = 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution