2· For acid solutions comprised of a moderately concentrated mixture of strong a

Question

2· For acid solutions comprised of a moderately concentrated mixture of strong and weak acids, the H concentration is approximately equal to the n of the strong acid (eg. acid (eg HNO»). This approximation a applies to the carry out. In your experiment, acid rain T acid rain experiment you are about to samples will involve the following reactions: weak acid : HNO, -H+ + NO strong acid: HNO,>H NO From these reactions, you can recognize that the strong acid, nitric acid, completely dissociates into hydrogen ions and nitrate ions, whereas the weak acid, nitrous acid, is predominately in its molecular form, HNO2 A. One simple way to measure the H is to measure pH. Ifa 10.0 ml solution of "acid rain" is measured to have a pH=2.77, what is the concentration of strong acid, HNO, formed? I lint: pl 1=-kg[I 11 and 10-11-11 r], where [HI is in molarity units. B. The total acid concentration (nitric +nitrous acid) of the 10.0 ml solution of “acid rain" is determined bv an acid-base titration method. You add 5.00 mL of the titrant sodium hydroxide with a concentration of 0.0114 M to reach the endpoint. i. What is the number of moles of hydroxide ions added? i. Write one net ionic equation for the acid-base reaction occurring during titration (Ihink in context of what chemical species in solution are reacting with each other to get a color change at endpoint). ii. What is the total number of moles of hydrogen ions in the "acid rain" sample? iv. What is the total concentration of hydrogen ions in the "acid rain" sample? v. What is the concentration of the weak acid, nitrous acid, in the solution of "acid rain" considered above (in 2-A)? Fall 2017 55B E M 1 1 10 FALL2 01 7 T O PIC 4 vi. Refer to the equilibrium reaction for the weak acid, HNO.. How will the equilibrium shift if the concentration of hydrogen ions decreases?

Explanation / Answer

a)

if pH = 2.77 thne...

[H+]= 10^-pH

[H+] = 10^-2.77

[H+] = 0.00169 M

b)

i)

mol of NAOH = MV = 0.0114*5*10^-3

mo of NaOH = 0.000057 mol

ii)

net ionic:

HNO2(aq) + OH-(aq) <->H2O(l) + NO2-(aq)

iii)

mol of base = mol of acid

then

0.000057 mol of base --> 0.000057mol of acid

iv)

[H+] = mol of H+ / Vtotal = 0.000057 / (10*10^-3) = 0.0057 M

v)

find weak acid / HNO2 concentration

[HNO2] = [H+] = 0.0057 M approx

vi)

equilibirum is

HNO2 <->H+ + NO2-

the equilibrium shifts from HNO2 to more HJ+ and NO2- (right side)

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