Titration of 50.00 mL HNO, with 0 3850 M NaOH 12 10 10 15 20 25 30 mL NaOH Added
ID: 1017536 • Letter: T
Question
Titration of 50.00 mL HNO, with 0 3850 M NaOH 12 10 10 15 20 25 30 mL NaOH Added 1. a Determine the following from the above graph: Equivalence Point mL Initial Equivalence Point mL pH pH Write the reactions with the appropriate arrows: The HNO2 with NaOH reaction equation: The salt dissociation equation: The salt hydrolysis equation: What is the molarity of the HNO, solution? (Use the volume and molarity of the base) b. c. What is the molarity of the HNO2 solution? d. What is the value of K, for HN02 calculated from the ½ equivalence point?Explanation / Answer
Titration plot
1.
a. Table
Initial : pH = 2.1 ; [H+] = 7.94 x 10^-3 M
Equivalence point : ml = 18.5 ml ; pH = 7.98
1/2 equivalence point : ml = 9.25 ; pH = 3.1
b. Reactions,
When HNO2 + NaOH,
HNO2 + NaOH --> NaNO2 + H2O
Salt dissociation : NaNO2 ---> Na+ + NO2-
Salt hydrolyzed ; NaNO2 + H2O <==> HNO2 + OH-
c. Molarity of HNO2 solution = 0.3850 M x 18.5 ml/50 ml = 0.142 M
d. Ka for HNO2 = 7.94 x 10^-4 [by graph 1/2 eq. point]
e. Ka by molarity and initial pH
Ka = (7.94 x 10^-4)^2/0.104 = 6.56 x 10^-4
f. pH at Equivalence point is dependent upon NaNO2 concentration
g. pH by ICE table
NaNO2 + H2O <===> HNO2 + OH-
initial 0.104 - -
change -x +x +x
Eq 0.104-x x x
So,
Ka = 6.56 x 10^-4 = x^2/0.104-x
x is small value can be ignored in denominator
x = [OH-] = 8.26 x 10^-3 M
[H+] = 1 x 10^-14/8.26 x 10^-3 = 1.21 x 10^-12 M
pH = -log[H+] = 11.91
h. At 25 ml NaOH, pH is dependent upon OH-
i. pH at 25 ml NaOH addition
excess [OH-] = 0.3850 M x (25 - 18.50) ml/75 ml = 0.033 M
pOH = -log[OH-] =1.47
pH = 14 - pOH = 12.52
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