Potassium hydrogen phthalate, KHC 8 H 4 O 4 , is used to standardize solutions o
ID: 564971 • Letter: P
Question
Potassium hydrogen phthalate, KHC8H4O4, is used to standardize solutions of bases. The acidic anion reacts with strong bases according to the following net ionic equation:
If a 0.781-g sample of potassium hydrogen phthalate is dissolved in water and titrated to the equivalence point with 47.00 mL of NaOH(aq), what is the molar concentration of the NaOH?
M
If 50.00 mL of the NaOH(aq) solution is added to an empty beaker and titrated to the equivalence point with 40.17 mL of HI(aq), what is the molar concentration of the HI?
M
Explanation / Answer
KHC8H4O4 (aq) + NaOH (aq)---------------> KNaC8H4O4(aq) + H2O(l)
K^+ (aq) +HC8H4O4^- (aq) + Na^+ (aq) +OH^- (aq)---------------> K^+(aq) +Na^+ (aq) +C8H4O4^2-(aq) + H2O(l)
removal of spectator ions to get net ionic equation
HC8H4O4^- (aq) + OH^- (aq)---------------> C8H4O4^2-(aq) + H2O(l)
no of moles of KHC8H4O4 = W/G.M.Wt
= 0.781/204.22 = 0.0038 moles
1 moles of KHC8H4O4 react with 1 moles of NaOH
0.0038 moles of KHC8H4O4 react with 0.0038 moles of NaOH
no of moles of NaOH = molarity * volume in L
0.0038 = molarity * 0.047
molarity = 0.0038/0.047 = 0.081 M
molarity of NaOH = 0.081M
2. NaOH(aq) + HI(aq) ----------------> NaI(aq) + H2O(l)
1 mole 1moles
NaOH is standard solution . let molarity of NaOH = 1M
NaOH HI
M1 = 1 M2 =
V1 = 50ml V2 = 40.17ml
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
M2 = M1V1n2/n1V2
= 1*50*1/1*40.17 = 1.24M
The molar concentration of HI = 1.24M
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