Problem:
A student titrates an unknown solution of H2SO4 with a standard solution of NaOH. The following data are obtained:
Initial reading of acid burette: 0.15 mL
Initial reading if base burette: 1.05 mL
Final reading of acid burette: 22.25 mL
Final reading of base burette: 46.50 mL
Concentration of standard base solution: 0.558 M
Calculate the concentration of the unknown acid solution. Hint: Think carefully about the equation for the reaction before you do the calculations.
A student titrates an unknown solution of H2SO4 with a standard solution of NaOH. The following data are obtained:
Initial reading of acid burette: 0.15 mL
Initial reading if base burette: 1.05 mL
Final reading of acid burette: 22.25 mL
Final reading of base burette: 46.50 mL
Concentration of standard base solution: 0.558 M
Calculate the concentration of the unknown acid solution. Hint: Think carefully about the equation for the reaction before you do the calculations.
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H2SO4 + 2NaOH ----> Na2SO4 + 2H2O
moles of base = 0.558*(46.5 - 1.05)*10^-3 = 0.02536 mol
It takes 2 moles of base to neutralize 1 mole of acid, so
moles of acid = ½ moles of base = 0.01268 mol
This no of moles was in (22.5 - 0.15)*10^-3 L or 22.35*10^-3 L of acid
Acid concentration = 0.01268/(22.35*10^-3) mol/L = 0.567 M
moles of base = 0.558*(46.5 - 1.05)*10^-3 = 0.02536 mol
It takes 2 moles of base to neutralize 1 mole of acid, so
moles of acid = ½ moles of base = 0.01268 mol
This no of moles was in (22.5 - 0.15)*10^-3 L or 22.35*10^-3 L of acid
Acid concentration = 0.01268/(22.35*10^-3) mol/L = 0.567 M
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That's so easy.. are you retarded or something?