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?
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"
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.
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