This question concerns the titration of a solution of sodium hydroxide with a solution of hydrochloric acid. As the titration proceeds the pH of the mixture changes. (a) What was the pH when 24.95 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to 25 cm3 of 1.00 mol dm–3 HCl(aq)? A 3 B 6 C 8 D 11
(b) What was the pH when 25.05 cm3 of 1.00 mol dm–3 NaOH(aq) had been added to 25 cm3 of 1.00 mol dm–3 HCl(aq)? A 3 B 6 C 8 D 11
i can calculate part (a) but i cant calculate (b) by using the same method. someone please help!
With part a) you end up with excess protons in solution after the reaction has taken place. But with part b) [OH-]>[H+] so you end up with hydroxide ions in solution after the reaction has taken place.
Does that help?
You need to find the hydroxide concentration and then use Kw=[H+(aq)][OH−(aq)] or 14=pOH + pH.
With part a) you end up with excess protons in solution after the reaction has taken place. But with part b) [OH-]>[H+] so you end up with hydroxide ions in solution after the reaction has taken place.
Does that help?
You need to find the hydroxide concentration and then use Kw=[H+(aq)][OH−(aq)] or 14=pOH + pH.
oh, i got it. thz i have one more question to ask,which say
This question concerns four solutions, A to D. They were prepared by mixing equal volumes of 0.2 mol dm–3 solutions of two different substances. The substances were A HCl(aq) and NaOH(aq) B HCl(aq) and NaCl(aq) C NH3(aq) and NH4Cl(aq) D CH3COOH(aq) and CH3CO2Na(aq) Select, from A to D, the mixture which would have a chloride ion concentration of 0.2 mol dm–3.
oh, i got it. thz i have one more question to ask,which say
This question concerns four solutions, A to D. They were prepared by mixing equal volumes of 0.2 mol dm–3 solutions of two different substances. The substances were A HCl(aq) and NaOH(aq) B HCl(aq) and NaCl(aq) C NH3(aq) and NH4Cl(aq) D CH3COOH(aq) and CH3CO2Na(aq) Select, from A to D, the mixture which would have a chloride ion concentration of 0.2 mol dm–3.
Write out a balanced equation for each reaction and work out the stoichiometric ratios