Given the following reaction: 2CrO4^2-(aq) + 2H^+(aq) <---> Cr2O7^2-(aq)+H2O(l)

Question

Given the following reaction: 2CrO4^2-(aq) + 2H^+(aq) <--->

Cr2O7^2-(aq)+H2O(l) Yellow orange

a. What color would a K2CrO4

solution be?

b. If sulfuric acid (H2SO4) is added to this solution,

will a color change be observed? If so, how does the addition of

sulfuric acid result in a color change? Explain your reasoning by

showing the effect of the addition of H2SO4 on the equilibrium for

the reaction.

c. If sodium hydroxide (NaOH) is added to the

solution, will a color change be observed? If so, how does the

addition of sodium hydroxide result in a color change? Explain your

reasoning by showing the effect of the addition of NaOH on the

equilibrium for the reaction.

Explanation / Answer

(a) Color of K2CrO4 solution = color of CrO4^2- = yellow


(b) 2 CrO4^2-(aq) + 2 H^+(aq) <=> Cr2O7^2-(aq) + H2O(l)

H2SO4(aq) => 2 H^+(aq) + SO4^2-(aq)


Adding H2SO4 => adding H^+

=> equlibrium shifts to right according to Le Chatelier's principle

=> Cr2O7^2- is formed

=> color will be orange


(c) 2 CrO4^2-(aq) + 2 H^+(aq) <=> Cr2O7^2-(aq) + H2O(l)

NaOH(aq) => Na^+(aq) + OH^-(aq)

OH^-(aq) + H^+(aq) => H2O(l)


Adding NaOH => adding OH^- => removing H^+

=> equlibrium shifts to left according to Le Chatelier's principle

=> CrO4^2- is formed

=> color will be yellow

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