1. What is the pH at the equivalence point in the titration of a 20.5 mL sample of a 0.437 M aqueous hydrocyanic acid solution with a 0.435 M aqueous potassium hydroxide solution?
pH =
2. When a 15.4 mL sample of a
0.408 M aqueous nitrous acid
solution is titrated with a 0.322 M aqueous
sodium hydroxidesolution, what is the pH at the
midpoint in the titration?
pH =
1)
HCN + KOH ----------> KCN + H2O
initially millimoles of HCN = 20.5 x 0.437 = 8.96
8.96 millimoles of KOH must be added to reach equivalence point .
8.96 = V x 0.435
V = 20.60 mL KOH must be added
total volume = 20.5 + 20.6 = 41.1 mL
[KCN] = 8.96 / 41.1 = 0.218 M
KCN is salt of weak acid for such salts
pH = 1/2 [pKw + pKa + log C]
pKa of HCN = 9.2 standard value
pH = 1/2 [14 + 9.2 + log 0.218]
pH = 1/2 [22.54]
pH = 11.27
pH = 11.3
2) HNO2 + NaOH ------------> NaNO2 + H2O
titration mid point means 50 % of reaction over
means resulting dolution contain 50% HNO2 and 50 % NaNO2
means [HNO2] = [NaNO2]
in such case for any titration
pH = pKa
pKa of HNO2 = 3.15
pH = 3.15
1. What is the pH at the equivalence point in the titration of a 20.5 mL sample of a 0.437 M aqueous hydrocyanic acid so...
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