Question

4. Suppose a sample of acid had a measured pH of 1.21, calculate the concentration of the acid. 5. Another solid acid that can be used as a primary standard is oxalic acid, which usually exists in the dihydrate form: H2C2O42H20. 2NaOH (ag) + H2C2042H20 (ag) Na2C204 (ag)+4H20 (1) If 0.231 g of oxalic acid dihydrate is neutralized by 31.75 mL of NaOH, calculate the molarity of the NaOH.

Answer 1

4) The concentration of acid can be calculated using following equation,

[H⁺] = 10^-pH

Here, pH = 1.12

Therefore,    [H⁺] = 10^-1.12 = 7.585 x 10^-2 M

Thus the pH of acid will be 7.585 x 10^-2 M

5) Let us calculate the number of moles of oxalic acid dihydrate present in 2.31 g by using following formula,

Number of moles of Oxalic acid dihydrate = Weight of oxalic acid dihydrate / molecular weight

The molecular weight of oxalic acid dihydrate is 126 g/mol

number of moles of oxalic acid dihydrate = 2.31 g / 126 g/mol

= 0.0183 mol

As from the balanced chemical equation, it can be seen that 1 mole of Oxalic acid dihydrate requires 2 moles of NaOH. Therefore,

number of moles of NaOH needed to nuetralize 0.0183 moles of oxalic acid dihydrate = 2 x number of moles of oxalic acid dihydrate

= 2 x 0.0183 = 0.0366 moles

Therefore, 31.75 mL of solution must be containing 0.0366 moles of NaOH. Now let us calculate the number of moles of NaOH in 100 mL

As, 31.75 mL = 0.0366 moles

1000 mL = (1000 x 0.0366) / 31.75 = 1.15 moles

Therefore, the molarity of NaOH solution will be 1.15 M