Question

Why does the NaOH solution have to be standardized?

The stockroom will sparge (displace the CO₂ with N₂ gas) the distilled water that we use for the preparation of the standard NaOH solution in order to remove excess CO₂ from the solution. Why must the CO₂ be removed? (Hint: Think about what happens when CO₂ reacts with H₂O.)

Calculate the concentration (in mol/L) of 30% by weight (30 g NaOH per 100 g of solution) NaOH solution. (The density of the 30% by weight solution is 1.35 g/mL). Show your work.

What volume of this 30% by weight solution is needed to make 250 mL of a ~0.100 M solution? Show your work.

Potassium hydrogen phthalate, KHC₈H₄O₄ (molar mass = 204.22g/mol), is a primary standard acid that reacts with sodium hydroxide on a 1:1 molar basis. A sample of the acid weighting 0.5893 g was titrated with NaOH, requiring 22.49 mL to reach the end point. Calculate the molar concentration of the NaOH solution.

Determine the pH of 100.0 mL of a 0.100 M H₃PO₄ solution.

Imagine you are titrating a 25.00 mL aliquot of ~0.100 M H₃PO₄. Assume that the titrant is 0.100 M NaOH. Use the example in the experiment background to help you calculate the volume of NaOH necessary to reach the first equivalence point. Calculate the pH of the solution at this point.

Repeat the calculations in problem 7 above for the second equivalence point.

Identify the species present at the 1^st, 2^nd, and 3^rd equivalence points.

Calculate the volume range necessary to titrate an unknown acid sample to the first equivalence point. (Refer to the protocol below for the mass used in the experiment and use the molar masses provided at the end of this document.)

Answer 1

Why does the NaOH solution have to be standardized?

There are two reasons for this.

1) NaOH is hygroscopic (absorbs water from atmosphere), So it can not be weighed accurately.

2) NaOH solution readily takes CO₂ from atmosphere. CO₂ dissolves in water to give weak acid(H₂CO₃, carbonic acid). This undergoes neutralization reaction with NaOH and decreases the concentrationof NaOH in the solution.

Why must the CO2 be removed?

The answer is same as above point 2.

Calculate the concentration (in mol/L) of 30% by weight (30 g NaOH per 100 g of solution) NaOH solution. (The density of the 30% by weight solution is 1.35 g/mL).

Molarity = number of moles / volume in liters

Number of moles = weight /molecular weight = 1.35 g /40 =0.03375 moles

So, concentration in molarity = 0.03375 moles/ 0.001 liters = 33.75 moles/liter

What volume of this 30% by weight solution is needed to make 250 mL of a ~0.100 M solution? Show your work.

Concentration of 30% by weight soltution (calculated above) = 33.75 M

Now using V_intialM_intial = V_Final M_Final​ ;

V_{intial *33.75 M = 250 * 0.1M}

V_intial = 0.74 mL is needed

Potassium hydrogen phthalate, KHC8H4O4 (molar mass = 204.22g/mol), is a primary standard acid that reacts with sodium hydroxide on a 1:1 molar basis. A sample of the acid weighting 0.5893 g was titrated with NaOH, requiring 22.49 mL to reach the end point. Calculate the molar concentration of the NaOH solution.

First lets calculate the moles of Potassium hydrogen phthalate = 0.5893 / 204.22 = 0.00289 moles

Since reaction is 1:1, at the end point 0.00289 moles of Potassium hydrogen phthalate reacts with  0.00289 moles of NaOH.

It took 22.49 mL, means, this much volume contains 0.00289 moles of NaOH

Concentration = number of moles / volume in liters = 0.00289 mol /0.02249 L= 0.12850 M