s) Predict the pH of the following and place an \"X\" in the appropriate box. (2
ID: 697697 • Letter: S
Question
s) Predict the pH of the following and place an "X" in the appropriate box. (2 points each) pH 7.00 NH, (k-1.8 × 10-5) is titrated with HCl to the equivalence point A) pH7.00. Not enough information given B) 50.0 mL of 0.200 M HNO, is reacted with 100 mL of 0.100 M NaOH Not enough pH 7.00 information given C) HC2H:O2 (K-1.8x 10) is titrated with NaOH to the equivalence point pH 47.00 PH-7,00 PH 700information given D) 0.400 M HCl is added to 0.100 M KOH Not enough -7.00 l pH > 7.00 | inform ationgiven E) 25.0 mL of 0.100 M NaOH is added to 25.0 mL of 0.200 M KOH pH 7.00 Not enough information given F) 0.150 M HC H,02 (K,-1.8 x 10*) is added to 0.200 M HCIExplanation / Answer
A) At equivalence point;
NH3 + HCl ---> NH4+ + Cl-
Ammonia reacted completely and forms NH4+ ion. This ion again in equilibrium with H2O
as follows:
NH4+ + H2O ----> NH3 + H3O+
we have Kb of NH3 = 1.8 x 10-5
Therefore,
Ka = Kw/Kb = 10-14/ 1.8x 10-5 = 5.6 x 10-10
NH4+ + H2O ----> NH3 + H3O+
Initial I - 0 0
change -x - +x +x
equilibrium I-x - x x
ka = x2/(I-x)
5.6 x 10-10 x I = x2
x = sqroot(5.6 x 10-10 x I)
x < 10-7
Since, x = [H3O+] < 10-7
The pH < 7.
B) HNO3 + NaOH ----> NaNO3 + H2O
1 mole fo HNO3 reacts with 1 mol of NaOH;
no. of moles of HNO3 = molarity x volume in L
= 0.2 mol/L x 0.05 L
= 0.01 mol
No. of moles of NaOH = molarity x volume in L
= 0.1 mol/L x 0.1 L
= 0.01 mol
Therefore equal moles of HNO3 and NaOH reacted and it is a complete neutralization reaction.
Hence, the pH = 7
C) This exactly opposite to the example (A) and the answer is pH >7.
D) Since, we do not have how much volume of HCL and KOH is mixed, we can't estimate the
pH. Therefore, not enough information given is the correct answer.
E) In this case, we NaOH and KOH solutions are mixed together. Both, these solutions are
bases, hence the resulting solution is also base. Hence, the pH will be greater than 7.
Answer: pH>7
F) HC2H3O2 is a weak acid and we are adding a concentrated solution of strong acid. Hence,
the resulting solution will be acid. Therefore, the solution pH <7
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