Calculate the pH of a 0.95 g in 1 L solution of nitrous acid, HNO2 at 25 degrees

Question

Calculate the pH of a 0.95 g in 1 L solution of nitrous acid, HNO2 at 25 degrees celsius. **PLEASE ANSWER 2 AND 3 WITH ALL WORK** thank you!!

2. Calculate the pH of a 0.95g in 1L solution of nitrous acid, HNO2 at 25 9C. (3pts) Ka(HNO2) 4.5 x 104 (Hint: Using the quadratic formula) 3. Calculate the equilibrium constant for OBr if a 0.1M solution of NaoBr has a pH of 10.85. (2pts) Write the ionic equation of salt hydrolysis. (1pts) Based on this equation, classify the salt as acidic, basic or neutral. (1 pts)

Explanation / Answer

2)

Molar mass of HNO2,

MM = 1*MM(H) + 1*MM(N) + 2*MM(O)

= 1*1.008 + 1*14.01 + 2*16.0

= 47.02 g/mol

mass,m = 0.95 g

number of mol,

n = mass/molar mass

= 0.95/47.02

= 0.0202 mol

volume , V = 1 L

Molarity,

M = number of mol / volume in L

= 0.02/1.0

= 0.0202 M

for simplicity lets write weak acid HNO2 as HA

HA -----> H+ + A-

0.0202 0 0

0.0202-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

since ka is small, x will be small and it can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4.5*10^-4)*0.0202) = 3.015*10^-3

pH = -log [H+] = -log (3.015*10^-3) = 2.52

Answer: 2.52

I am allowed to answer only 1 question at a time

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