Question

13)

A 1.0-L buffer solution is 0.105 M in HNO2 and 0.175 M in NaNO2.

Part A

Determine the concentrations of HNO2 and NaNO2 after addition of 1.2 g HCl.

Express your answers using three significant figures separated by a comma.

[HNO2], [NaNO2] =		M

Part B

Determine the concentrations of HNO2 and NaNO2 after addition of 1.2 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 1.2 g HI.

Express your answers using three significant figures separated by a comma.

[HNO2], [NaNO2] =		M

Answer 1

PART A

[HNO2] = 0.138M

[NaNO2] = 0.142M

Explanation

Initial moles of HNO2 = 0.105mol

Initial moles of NaNO2 = 0.175mol

Number of moles = mass/molar mass

Number of moles of HCl added = 1.2g/36.46g/mol = 0.03291mol

HCl react with NaNO2

NaNO2(aq) + HCl(aq) -----> HNO2(aq) + NaCl(aq)

0.03291 moles of HCl reacts with 0.03291moles of NaNO2 to give 0.03291moles of HNO2

After addition of HCl

Number of moles of HNO2 = 0.105mol + 0.03291mol= 0.1379mol

Number of moles of NaNO2 = 0.175mol - 0.03291mol = 0.1421mol

concentration of HNO2 = 0.138 M

concentration of NaNO2 = 0.142 M

PART B

[HNO2] = 0.0750M

[NaNO2] = 0.205M

Explanation

Number of moles of NaOH added = 1.2g /40g/mol = 0.03mol

NaOH reacts with HNO2

HNO2(aq) + NaOH(aq) ------> NaNO2(aq) + H2O(l)

0.03moles of NaOH reacts with 0.03moles of HNO2 to produce 0.03moles of NaNO2

After addition of NaOH

Number of moles of HNO2 = 0.105mol - 0.03mol = 0.075mol

Number of moles of NaNO2 = 0.175mol + 0.03mol = 0.205mol

[HNO2] = 0.0750M

[NaNO2] = 0.205M

PART C

[HNO2] = 0.114M

[NaNO2] = 0.166M

Explanation

Number of moles of HI added = 1.2g/ 127.91g/mol = 0.0093816mol

HI reacts with NaNO2

NaNO2(aq) + HI(aq) ------> NaI(aq) + HNO2(aq)

0.0093816moles of HI reacts with 0.0093816 moles of NaNO2 to give 0.0093816moles of HNO2

After addition of HI

number of moles of HNO2 = 0.105mol + 0.0093816mol = 0.11438mol

number of moles of NaNO2 = 0.175mol - 0.0093816mol = 0.16562mol

[HNO2] = 0.114M

[NaNO2] = 0.166M