25.00 mL of HNO_3 (nitric acid) sample was titrated with 0.500 M of NaOH. It was

Question

25.00 mL of HNO_3 (nitric acid) sample was titrated with 0.500 M of NaOH. It was found that phenolphthalein indicator added into nitric acid sample changed from colorless to pale pink after adding 12.00 mL of NaOH. Write the net ionic equation for this titration. Calculate the concentration of the nitric acid sample. .. M How many mL of NaOH was added when the titration was at the half equivalence point? ... mL Calculate pH at half-equivalence point. PH = ... Calculate pH at the equivalence point (approximately the same as the end point). PH = ...

Explanation / Answer

Q2.

a)

Net ionic equation for HNO3 + NaOH:

HNO3(aq) + NaOH(aq) --> H2O(l) + NaNO3(aq)

ionic

H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) --> H2O(l) + Na+(aq) + NO3-(aq)

net ionic includes only those ions reacitng

H+(aq) + OH-(aq) --> H2O(l)

B)

the concentration of acid:

mol of acid = mol of base

mol of base = MV = (0.5)(12*10^-3) = 0.006 mol of base

so

mol of acid = 0.006 mol

M acid = mol/V = 0.006/(25*10^-3) = 0.24 M of acid

c)

NaOH required for half equivalence point, by definition, it is the HALF of that used in equivalence so:

12 mL --< 1/2¨*12 = 6 mL

d)

pH in half equivalnece point

mmol of acid initially = 0.006 *10^3 = 6

mmol of base added = MV = 0.5*6 = 3

so..

mmol of acid left = 6-3 = 3

V total = 25+6 = 31 mL

[H+] = mmol/mL = 3/31 = 0.096774

pH = -log(H) = -log(0.096774 = 1.01424

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