Answer to Question #175300 in General Chemistry for mel

Question

Determination of Percent KHP in a mixture by Acid/Base Titration

Standardization of NaOH:

Ttrial 1:

• mass of KHP = 0.500 g
• buret reading (final) = 25.00 mL
• buret reading (initial) = 0 mL
• NaOH used = ?
• molarity of NaOH: ?M

Trial 2:

• mass of KHP = 0.500 g
• buret reading (final) = 25.40 mL
• buret reading (initial) = 0 mL
• NaOH used = ?
• molarity of NaOH: ?M

Trial 3:

• mass of KHP = 0.500 g
• buret reading (final) = 24.80 mL
• buret reading (initial) = 0 mL
• NaOH used = ?
• molarity of NaOH: ?M

Average molarity from all 3 trials: ?

Determination of % KHP in unknown sample:

Trail 1:

• mass of sample: 0.600 g
• buret reading (final) = 16.00 mL
• buret reading (initial) = 0
• NaOH used = ?

Trail 2:

• mass of sample: 0.600 g
• buret reading (final) = 15.50 mL
• buret reading (initial) = 0
• NaOH used = ?

Trail 3:

• mass of sample: 0.600 g
• buret reading (final) = 15.80 mL
• buret reading (initial) = 0
• NaOH used = ?

calculation of mass of KHP and % KHP = ?

Answer 1

$Molarity=mass/rfm$

Trial 1= 25/40=0.625M

Trial 2= 25.4/40= 0.635M

Trial 3 = 24.8/40=0.62M

Average molarity= 0.63M