A buffer solution is prepared by add 175.5 mL of 0.155 M Phenol HC_6H_5O, to 100
ID: 1063603 • Letter: A
Question
A buffer solution is prepared by add 175.5 mL of 0.155 M Phenol HC_6H_5O, to 100 mL of 0.0855 M lithium phenolate. LiC_6H_5O. Part 1 What is the pH of this buffer solution? Part2: What is the pH of this buffer solution if 85 mL of 0.088 M NaOH is added? (Phenol: K_a = 1.3 times 10^-10) 9.38; 9.80 10.11; 10.78 9.38; 8.97 9.80; 9.38 8.97; 9.38 A nitric acid solution. HNO_3, will be titrated with potassium hydroxide. KOH. What is the pH of a 65 mL sample of a 0.15 M HNO_3 solution? What is the pH of the solution after adding 100 mL of 0.095 M KOH to the 65 mL of 0.15 M HNO_3 solution? pH = 0.471; pH = 2.02 pH = 2.41; pH = 3.05 pH = 0.823; pH = 1.13 pH = 1.53; pH = 3.17 pH = 0.823; pH = 2.82 A nitric acid solution, HNO_3, will be titrated with potassium hydroxide, KOH. What is the pH of the solution after adding 102.6 mL of 0.095 M KOH to the 65 mL of 0.15 M HNO_3 solution? What is the pH of the solution after adding 105.5 mL of 0.095 M KOH to the 65 mL of 0.15 M HNO_3 solution? pH = 7.00 (Equivalence Point); pH = 2.79 pH = 5.11: pH = 6.73 pH = 6.05; pH = 12.07 pH = 7.00 (Equivalence Point); pH = 11.20 pH = 7.00 (Equivalence Point); pH = 9.62 A formic acid solution. HCHO_2, will be titrated with a sodium hydroxide, NaOH, solution. What is the pH of an 85 mL sample of a 0.125 M HCHO_2 solution? What is the pH of the solution after adding 53 mL of 0.1002 M NaOH to the 85 mL of 0.125 M HCHO_2 solution? (Formic Acid: K_a = 1.7 times 10^-4)Explanation / Answer
Q15
V = 175.5 mL M = 0.155 M phenol
V = 100 mL M = 0.0855 M phenolate
find pH of buffer....
a)
Ka = 1.3*10^-10
pKa = -log(Ka) = -log(1.3*10^-10) = 9.88
for a buffer,
apply Hendersno hasselbach equation
pH = pKa + loG(A-/HA)
pH = 9.88 + log(A/HA)
HA = M1V1/(V1+V2) = 175.5*0.155 / (175.5+100) = 0.09873
A- = M2V2/(V1+V2) = 100*0.0855/ (175.5+100) = 0.0310344
pH = 9.88 + log(A/HA) = 9.88 + log(0.0310344/0.09873) = 9.37739
b)
after
V = 85 mL of M = 0.088 M of NaOH
mmol of base = MV = 85*0.088 = 7.48 mmol of OH-
so..
mmol of acid = MV = 175.5*0.155 = 27.2025
mmol of conjguate = MV = 100*0.085 = 8.5
after 7.48 mol of OH- addition
acid left =27.2025-7.48 = 19.7225
conjguate formed = 8.5+7.48 = 15.98
so
pH = 9.88 + log(A/HA) = 9.88 + log(15.98/19.7225) = 9.78
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