The final volume of buffer solution must be 100.00 mL and the final concentratio

Question

The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M.

Based on this information, what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH=7.0 (in grams)? Based on this information, what volume of acid should the student measure to make the 0.100 M buffer solution (in mL)?

My weak acid is Sodium dihydrogen phosphate monohydrate, Ka= 6.23X10^-8, 2.00M and my conjugate base is Disodium hydrogen phosphate heptahydrate, Na2HPO4*7H2O

Explanation / Answer

Sol:-

Given final concentration of weak acid i.e [NaH2PO4.H2O] = 0.100 M

Also final volume = 100.00 mL = 0.100 L

Ka of NaH2PO4.H2O = 6.23 x 10-8

therefore

pKa = - log Ka = - log 6.23 x 10-8 = - ( - 7.20) = 7.20

so

pKa of NaH2PO4.H2O = 7.20 .

given that pH of acidic buffer solution = 7.0

Now by using Henderson-Hasselbalch equation , we have

pH = pKa + log [conjugate base ] / [ Acid ]    

pH = pKa + log [ Na2HPO4.7H2O] / [ NaH2PO4.H2O]

7.0 = 7.20 + log [ Na2HPO4.7H2O] / [ NaH2PO4.H2O]

log [ Na2HPO4.7H2O] / [ NaH2PO4.H2O] = 7.0 - 7.20

log [ Na2HPO4.7H2O] / [ NaH2PO4.H2O] = -0.20

[ Na2HPO4.7H2O] / [ NaH2PO4.H2O] = 10-0.20

[ Na2HPO4.7H2O] / [ NaH2PO4.H2O] = 0.63

[ Na2HPO4.7H2O] = 0.63 x [ NaH2PO4.H2O]

[ Na2HPO4.7H2O] = 0.63 x 0.100 M

[ Na2HPO4.7H2O] = 0.063 M

Number of moles of Na2HPO4.7H2O / final volume in litre = 0.063 M
( because Molarity = number of moles of solute / volume of solution in litre )

therefore

Number of moles of Na2HPO4.7H2O = 0.063 M x 0.100 L

Number of moles of Na2HPO4.7H2O = 0.0063 mol

Also Number of moles of substance = mass of substance in gram / gram molar mass of substance .

and gram molar mass of Na2HPO4.7H2O = 268 g/mol

therefore

mass of Na2HPO4.7H2O = moles of Na2HPO4.7H2O x gram molar mass of Na2HPO4.7H2O

mass of Na2HPO4.7H2O = 0.0063 mol x 268 g/mol

mass of Na2HPO4.7H2O = 1.69 g

Hence mass of Na2HPO4.7H2O used = 1.69 g

Number of moles of weak acid i.e NaH2PO4.H2O = Molarity of NaH2PO4.H2O x final volume in litre

Number of moles of weak acid i.e NaH2PO4.H2O = 0.100 M x 0.100 L = 0.01 mol

given initial concentration of NaH2PO4.H2O = 2.00 M

Molarity = number of moles of solute / volume of solution in litre

therefore

Volume of solution = number of moles of NaH2PO4.H2O / Molarity of NaH2PO4.H2O

Volume = 0.01 mol / 2.00 mol/L

Volume = 0.005 L

Volume = 5.00 mL

Hence Volume of NaH2PO4.H2O used = 5.00 mL

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