Titration

20.0cm3 of a solution of NaoH containing 0.05mol of alkali in 500cm3 of solution requires 19.8cm3 of dibasic acid H2A to neutralize it.
1. Calculate the molarity of the acid
2. If the concentration of the acid is 4.90gdm3, what is the relative molecular mass of the acid to the nearest whole number
3.calculate the percentage by mass of A in H2A.

20.0cm3 of a solution of NaoH containing 0.05mol of alkali in 500cm3 of solution requires 19.8cm3 of dibasic acid H2A to neutralize it.
1. Calculate the molarity of the acid
2. If the concentration of the acid is 4.90gdm3, what is the relative molecular mass of the acid to the nearest whole number
3.calculate the percentage by mass of A in H2A.

Let me rephrase his question. He probably meant:
Since there are 0.05 mol of NaOH in 500 cm³ solution, the concentration = 0.05/ 0.5 dm³ = 0.1 mol dm⁻³. The solution was stored in a volumetric flask that has a capacity of 500 cm³. He then pipetted 20.0 cm³ aliquot of NaOH from that 500 cm³ volumetric flask and transferred it into a conical flask. As a result, the number of moles of NaOH pipetted = 20.0/ 1000 x 0.1 = 0.002 mol. Titration was subsequently performed and the dibasic acid was the titrant, while NaOH was the analyte.

1) 2 NaOH + H₂A → Na₂A + 2 H₂O
Based on the stoichiometric ratio, number of moles of H₂A = 0.002/ 2 = 0.001 mol.
Hence, molarity of H₂A = 0.001/ (19.8/1000) = 0.0505 mol dm⁻³. (corrected to 3 s.f)

2) mol dm⁻³ = g dm⁻³ / Mr
Hence, Mr = 4.90 / 0.0505 = 97.02
Relative molecular mass of acid = 97 (corrected to nearest whole number)
* In fact, it is possible that the dibasic is a sulfuric acid because its Mr = 98

3) Percentage by mass = (1x2)/ 97 x 100% = 2.06% (corrected to 3 s.f.)

Hope it helps. Cheers.

Please don't understand the way you solve number 3.i want the percentage by mass of A in the acid

"3) Percentage by mass = (1x2)/ 97 x 100% = 2.06% (corrected to 3 s.f.)" I "misread" the question.
This was for H. To get A, just subtract 2.06% from 100%.

Answer is 97.94% for percentage by mass of A in H₂A.

Hope it helps. Cheers.

"3) Percentage by mass = (1x2)/ 97 x 100% = 2.06% (corrected to 3 s.f.)" I "misread" the question.
This was for H. To get A, just subtract 2.06% from 100%.

Answer is 97.94% for percentage by mass of A in H₂A.

Hope it helps. Cheers.

Let me rephrase his question. He probably meant:
Since there are 0.05 mol of NaOH in 500 cm³ solution, the concentration = 0.05/ 0.5 dm³ = 0.1 mol dm⁻³. The solution was stored in a volumetric flask that has a capacity of 500 cm³. He then pipetted 20.0 cm³ aliquot of NaOH from that 500 cm³ volumetric flask and transferred it into a conical flask. As a result, the number of moles of NaOH pipetted = 20.0/ 1000 x 0.1 = 0.002 mol. Titration was subsequently performed and the dibasic acid was the titrant, while NaOH was the analyte.

1) 2 NaOH + H₂A → Na₂A + 2 H₂O
Based on the stoichiometric ratio, number of moles of H₂A = 0.002/ 2 = 0.001 mol.
Hence, molarity of H₂A = 0.001/ (19.8/1000) = 0.0505 mol dm⁻³. (corrected to 3 s.f)

2) mol dm⁻³ = g dm⁻³ / Mr
Hence, Mr = 4.90 / 0.0505 = 97.02
Relative molecular mass of acid = 97 (corrected to nearest whole number)
* In fact, it is possible that the dibasic is a sulfuric acid because its Mr = 98

3) Percentage by mass = (1x2)/ 97 x 100% = 2.06% (corrected to 3 s.f.)

Hope it helps. Cheers.