Question

The equivalent weight of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is?

Answer 1

The equivalent weight of K₂Cr₂O₇

An equivalent weight of a solution is defined as the molecular weight of the solute divided by the valence of the solute.

• Equivalent weight is used for predicting the mass of a substance that reacts with one atom of hydrogen in an acid-base analysis like in titration. Equivalent weight is used in the calculation of normality.
• Equivalent Weight
  Equivalent weight = molecular weight / X, where X is the valency factor ( number of charges on the positive ion)

Reaction

${\mathrm{K}}_2{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}\left(\mathrm{aq}\right)+14{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+6{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{K}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+3{\mathrm{I}}_2\left(\mathrm{g}\right)$

Step 1:calculating molecular weight of potassium dichromate

• The molecular weight of potassium dichromate $=(2\times \mathrm{atomic}\mathrm{weight}\mathrm{of}\mathrm{K})+(2\times \mathrm{atomic}\mathrm{weight}\mathrm{of}\mathrm{Cr})+(7\times \mathrm{atomic}\mathrm{weight}\mathrm{of}\mathrm{O})=(2\times 39)+(2\times 52)+(7\times 16)$

$=78+104+112=294\mathrm{amu}$

Step 2:calculating equivalent weight of potassium dichromate

• There is an overall gain of 6 electrons.
• That means the n-factor for this reaction is 6.
• molecular weight of potassium dichromate = 294 ,
• X factor in acidic medium = 6
• Equivalent weight = molecular weight / X factor

$=\frac{294}{6}=49$

Hence, the equivalent weight of potassium dichromate in an acidic medium is 49 g/eq.