Question

A 1.0-L buffer solution is 0.135 M in HNO2 and 0.195 M in NaNO2.

Part A Determine the concentrations of HNO2 and NaNO2 after addition of 1.3 g HCl. Express your answers using three significant figures separated by a comma. [HNO2], [NaNO2] =   M   SubmitMy AnswersGive Up Part B Determine the concentrations of HNO2 and NaNO2 after addition of 1.3 g NaOH. Express your answers using three significant figures separated by a comma. [HNO2], [NaNO2] =   M   Part C Determine the concentrations of HNO2 and NaNO2 after addition of 1.3 g HI. Express your answers using three significant figures separated by a comma. [HNO2], [NaNO2] =   M   SubmitMy AnswersGive Up

Answer 1

Initial conditions:

A 1.0-L buffer solution is 0.135 M in HNO2 and 0.195 M in NaNO2.

mol of HNO2 = MV = 0.135*1 = 0.135 mol HNO2

mol of NaNO2 = MV = 0.195*1 = 0.195 mol of NaNO2

PART A

Determine the concentrations of HNO2 and NaNO2 after addition of 1.3 g HCl.

mol of HCl = mass/MW = 1.3/36.4609 = 0.0357

initially:

0.135 mol HNO2

0.195 mol of NaNO2

after adding 0.0357 H+

0.135 mol HNO2 + 0.0357 H+ = 0.1707 mol of HNO2 --> 3 sig fig --> 0.171 mol of HNO2

0.195 mol of NaNO2- 0.0357 H+ = 0.1593 mol of NaNO2 --> 3sig fig --> 0.160 mol of NaNO2

Molarity = mol/V = mol/1 L

[HNO2] = 0.171 M

[NaNO2] = 0.160 M

b)

if we add 1.3 g of NaOH

mol of NAOH = mass/MW = 1.3 / 39.99 = 0.03250812 mol of NaOH

OH- reacts with HNO2 to form NaNO2

0.135 mol HNO2 - 0.03250812 H+ = 0.102491 mol of HNO2 --> 3 sig fig --> 0.102 mol of HNO2

0.195 mol of NaNO2 + 0.03250812 H+ = 0.2275 mol of NaNO2 --> 3sig fig --> 0.228  mol of NaNO2

[HNO2] = 0.102 M

[NaNO2] = 0.228 M

c)

this is similar to HCl,

mol of HI = mass/MW = 1.3/127.911 = 0.010163 mol

0.135 mol HNO2 + 0.010163 H+ = 0.145163 mol of HNO2 --> 3 sig fig --> 0.145 mol of HNO2

0.195 mol of NaNO2- 0.010163 H+ = 0.205163 mol of NaNO2 --> 3sig fig --> 0.205 mol of NaNO2

Molarity = mol/V = mol/1 L

[HNO2] = 0.145 M

[NaNO2] = 0.205 M