Concentration Calculations (1 Viewer)

[decimari]

Hello!
I need help with the following question:

35mL of 0.86 M sulfuric acid was placed into a 25mL solution of 1.00 M sodium hydroxide.

b) Calculate the concentration hydrogen or hydrogen ions after this reaction. You may assume the reaction is complete. (3 marks)

c) Calculate the pH of the final solution. (1 mark)

d) If 5 drops of methyl orange was placed into the NaOH solution before the reaction took place, explain the colour changes that occur over the course of the experiment. (3 marks)

Any help would be greatly appreciated!

 

[leehuan]

May need someone to check as it's been a while since I've done calculation questions.

H2SO4(aq) + 2 NaOH(aq) -> Na2SO4(aq) + 2 H2O(l)

Moles of H2SO4 present:
n=CV ---> n=0.86 * (35*10^-3) = 0.0301mol

Moles of NaOH present:
n=CV ---> n=1.00 * (25*10^-3) = 0.0250 mol

Hence NaOH is the limiting reagent and H2SO4 is in excess.

Moles of H2SO4 present AFTER reaction: 0.0301-0.0250 = 0.0051 mol
Combined volume after reaction: 35mL + 25mL = 0.060L

For every 1 mol of H2SO4 present, there exists 2 hydrogen (or more accurately hydronium) ions.
Therefore moles of H+ = 0.0102 mol

Concentration of the hydrogen ion
C=n/V
C=0.0102/0.060 = 0.17mol L^-1 (leave at 2s.f.)
____________________

pH = -log_10[H+] = -log_10(0.17) = 0.769551... = 0.77 (correct to 2s.f.)
____________________

Methyl orange changes colour over the acidic range. For strong acids, methyl orange will appear red, whereas for neutral and basic substances it will appear yellow instead.

If 5 drops of methyl orange was placed into the NaOH originally, then due to the highly basic nature of NaOH it will appear to be yellow. However, when the NaOH was reacted with the H2SO4, clearly the resulting solution became highly acidic, as determined by the pH calculated in c). Because the methyl orange is now in acidic solution, the indicator now turns red.