Question

A 1.0 L buffer solition is 0.115 M in HNO2 and 0.165 M in NaNO2.

CHW 7: Ch 17 Part 1 (Buffers, Titrations, Solubility & Complex lon)-3.8 hrs)(58 credits) Exercise 17.60 A 1.0-L buffer solution is 0.115 Min HNO, and 0.165 Min NaNO, Part A Determine the concentrations of HNO, and NaNO, after addition of 1.8 g HCL. Express your answers using three significant figures separated by a comma. Ad O ? [HNO.). (NaNOM Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining Part B Determine the concentrations of HNO, and NaNO, after addition of 1.8 g NaOH. Express your answers using three significant figures separated by a comma 180 AED O ? (HNO3). NaNO,) - Submit Rettu

please answer parts a,b, and c

Answer 1

moles of HNO2 = Molarity × volume

= 0.115 × 1.0

= 0.115

Moles of NaNO2

= 0.165 × 1.0

= 0.165

Part A.

When 1.8 g HCl is added

Moles of HCl = ( mass/molar mass)

= ( 1.8/36.5)

= 0.049

HCl + NaNO2 $\to$ HNO2 + NaCl

BCA table is

Moles	NaNO	HCl	HNO
Before	0.165	0.049	0.115
Change	-0.049	-0.049	+0.049
After	0.116	0	0.164

so, [HNO2] = ( moles/volume)

= (0.164/1) = 0.164 M

[NaNO2] = ( 0.116/1) = 0.116 M.

Part B.

Now when 1.8 g NaOH is added

Moles of NaOH = (1.8/40)

= 0.045

NaOH + HNO2 $\to$ NaNO2 + H2O

BCA table is

Moles	HNO2	NaOH	NaNO
Before	0.115	0.045	0.165
Change	-0.045	-0.045	+0.045
After	0.070	0	0.210

[HNO2] = (Moles/volume) = (0.070/1) = 0.070 M.

[NaNO2] = (0.210/1) = 0.210 M.

Part C.

When 1.8 g HI is added

Moles of HI

= ( Mass/molar mass)

= (1.8/128)

= 0.014

HI + NaNO2 $\to$ HNO2 + H2O

BCA table is

Moles	NaNO2	HI	HNO2
Before	0.165	0.014	0.115
Change	-0.014	-0.014	+0.014
After	0.151	0	0.129

Then,

[HNO2] = ( 0.129 mol / 1 L) = 0.129 M.

[NaNO2] = (0.151 mol /1 L ) = 0.151 M.