Question

If 16.8 mL of the analyte H2C2O4 (oxalic acid) is titrated with 0.501 M KOH to 13.54 mL, determine the mass of the analyte.​

Answer 1

Answer:

Mass of Analyte H2C2O4 = 18.18 grams

Explanation:

The balanced equation is:

$H_{2}C_{2}O_{4}+2KOH\rightarrow K_{2}C_{2})_{4}+H_{2}O$

The Molarity of Oxalic acid needed can be determined using:$M_{1}V_{1}=M_{2}V_{2}$

Here M1 = Molarity of Oxalic acid

V1 = Volume of oxalic acid

M2 = Molarity of KOH and V2 = Volume of KOH

M1 = ? , V1 = 16.8 mL , M2 = 0.501 M   ,  V2 = 13.54 mL

Put the values and solve for M2,

$M_{1}V_{1}=M_{2}V_{2}$

$2\times M_{1}(16.8)=0.501\times 13.54$

We are multiplying by 2 because , 1 H2C2O4 needs 2 KOH

$M_{1}=\frac{0.501\times 13.54}{2\times 16.8}$

$M_{1}=0.202M$

Hence Molarity of H2C2O4 =

0.202 M

Molar mass = H2C2O4 = 2x1 +2x12 + 4x16

= 2+24+64 =  90 gram

$Mole =molarity\times Molar\ mass$

$Moles=0.202\times 90$

18.18 gram