Answer to Question #92715 in Organic Chemistry for Lee

Question

Compare 1.0 mol/L sodium hydroxide with 1.0 mol/L ammonia (written as ammonium hydroxide) by recreating similar diagrams of dissociation and concentration as those provided to you for the acid comparison. Assume ammonia dissociates only 1%. What would you expect to see if you tested the conductivity and pH of each? Why is there a difference in reactivity if there is 1 mole of acid in a given litre?

Answer 1

1) $NaOH -> Na^++OH^-$

NaOH is a strong electrolyte, it dissociates completely. The concentration of $OH^-$ is the same as of NaOH, c($OH^-$ ) = 0.1 mol/L

$pH 14-pOH=14 -( -log_{10}0.1) = 14-1 = 13$

The solution of NaOH will be a good conductor of electricity as there are 0.1 mol of $Na^+$ ions and 0.1 mol of $OH^-$ions in 1 L of solution.

2) $NH_4OH\leftrightarrow NH_4^++OH^-$

NH4OH is a weak electrolyte. If only 1 % of NH4OH dissociates, then$c(NH_4^+) = 0.1\times0.01 = 0.001 mol/L$

$c(OH^-)=0.1\times0.01 = 0.001 mol/L$

$pH = 14-pOH = 14 -(-log_{10}0.001)=11$

The solution of NH4OH will be a bad conductor of electricity as the concentrations of both ions are very small.

Comparison of both solutions: the first solution contains more ions than the second one, therefore solution of NaOH will be a better conductor of electricity. Also pH of the first solution is bigger than that of the second one.

There is a difference in reactivity if there is 1 mole of acid in a given litre because the first solution contains a strong base and second solution contains a weak base.