Question

1) a) write balanced molecular and net ionic equations for the neutrilization of NaOH by KHP( mol. WT= 204.23). Khp is an abbreviation for potassium hydrogen phthalate, KHC8H4O4,a subtance of known high purity with one acid hydrogen as indicated by the formula.

b) if 2.8 g of 204.23 KHP is dissolved in water and titrated with 40 ml of NaOH to the phenolphthalein endpoint, what is the molarity of the NaOH solution?

c) this now standardization solution of NaOH is used to find the molecular weight of an acid H2A which has two acid hydrogen. write a balanced molecular equation for this titration reaction.

d) if 1.08 g of this acid requires 37.18 of the NaOH solution for titration to the appropriate endpoint. calculate the molecular weight of the unknown acid H2A.

Answer 1

a)

Balanced Molecular Equation:

NaOH(aq) + KHC8H4O4(aq) ---> H2O(l) + KNaC8H4O4(aq)

Net ionic equation

H+(aq) + OH-(aq) ---> H2O(l)

b)

If m = 2.8 g of KHP is dissolved in water

MW KHP = 204.23 g/mol

in V = 40 ml of NaOH

to endpoint

what is NaOH molarity?

Recall that

NaOH(aq) + KHC8H4O4(aq) ---> H2O(l) + KNaC8H4O4(aq)

Therefore 1 mol of NaOH needs one mol of KHP to neutralize

in the endpoint

moles of acid = moles of base

moles of acid = mass/MW = 2.8 g/ 204.23 = 0.01371 mol of KHP

therefore

0.01371 of Base

M = mol of Base / V

M = 0.01371 mol / 0.040 L = 0.3428 M

Concnetration of NaOH = 0.3428 M

c)

Balanced equation

2NaOH(aq) + H2A(aq) ---> 2H2O(l) + NaA(aq)

d)

m = 1.08 of acid

needs

37.18 ml of NaOH(assume 37.18 is ml an not moles)

Calculate moles of NaOH used

M = mol/V

mol = M*V = 0.3428M*37.18ml = 12.74 mmol of Base was used

since ratio is 2:1 (2 mol of base per mol of acids)

Then

12.74*(1/2) = 6.373 mmol of Acid is present

6.373 mmol --> 6.373*10^-3 mol

since we need MW

we can relate this as follows:

MW = mass/mol

MW = 1.08 g/ 6.373*10^-3 mol = 169.46 g/mol

MW of Acid = 169.46 g/mol