Chemistry, 16.03.2020 16:59 dlewis9632

A buffer is made by combining 5.55 g of NH3(Kb= 1.8 X 10-5)with 4.78 g of HCl and diluting to a volume of 750.0 mL. a. What is the pH?b. What is the pH after adding 60.0 mL of 2.00 M HBr(aq)?c. What is the pH after adding 120.0 mL of 2.00 M HBr(aq)?

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Answer from: ladybugys

a) pH = 9.41

b) pH after adding 60.0 mL of 2.00 M HBr(aq) = 9.06

c) pH after adding 120.0 mL of 2.00 M HBr(aq) = 8.98

Explanation:

Given that:

mass of NH_3 =5.55g

mass of HCl= 4.78 g

Total volume = 750.0 mL = 0.75 L

Equation for the reaction

NH_3+HCl ------NH_4Cl

The molar mass for NH_3 = 17 g/mol

molar mass of HCl = 36 g/mol

∴ number of moles = $\frac {mass} {molarmass}$

For NH_3 ;

number of moles = $\frac {5.55} {17}$

= 0.3265 mole of NH_3

For HCl ;

number of moles = $\frac {4.78} {36}$

= 0.1328 mole of HCl

The stoichiometric ratios for the reaction are in the ratio of 1:1

That implies that 1 mole of HCl reacts with 1 mole of NH_3

with HCl being the limiting reactant;

0.1328 mole of HCl react with 0.1328 mole of NH_3 to form 0.1328 mole of NH_4Cl

Final concentration of NH_3 = 0.3265 - 0.1328

= 0.1937 mole of NH_3

Final concentration of HCl = 0.1328 -0.1328

= 0.00 mole of HCl

However, concentration of NH_4Cl and NH_3 can be calculated as follows:

Since Concentration(mol/L) = $\frac{number of moles }{volume}$

For NH_3 ; we have:

= $\frac{0.1937}{0.75}$

= 0.2583 mole/L

For NH_4Cl ; we have :

= $\frac{0.1328}{0.75}$

= 0.1771 mole/L

Equation of the reaction in the solution of HBr is as follows:

NH_3 + HBr ----- NH_4^+ + Br^-

Dissociation of NH_4Cl yields:

NH_4Cl^- ----- NH^+_4 + Cl^-

Since, NH_4^+ is present in both of the above reactions; then this a buffer solution which is demonstrated by using Henderson Hasslelbach equation.

$pOH = pKb + log (\frac{HB^+}{HB} )$

given that Kb = $(1.8*10^{-5})$

pKb = -log (Kb)

pKb = $-log (1.8*10^{-5})$

pKb = 4.75

$pOH = pKb + log (\frac{HB^+}{HB} )$

$pOH = 4.75 + log (\frac{0.1771}{0.2583} )$

$pOH = 4.75+(-0.1639)\\pOH= 4.5861$

pH = 14 - pOH

pH = 14 - 4.5861

pH = 9.4139

pH = 9.41

b) After 2.00 M of HBr and Volume = 60.0 mL

Volume increases from 750mL to : (750 + 60 ) mL = 810 mL = 0.81 mL

Definitely the concentration of the Br⁻ will also increase.

The new concentration of [Br⁻] = $\frac{number of moles}{volume}$

moles of Br⁻ added = The new concentration of [Br⁻] × volume

= 60 ×2.0

= 120 mL. mol

mole of Br⁻ already in the solution = 0.1771 + $\frac{120}{1000}$

= 0.1771 + 0.12

= 0.2971 mol of Br⁻

[Br⁻] = $\frac{number of moles}{volume}$

[Br⁻] = $\frac{0.2971}{0.81}$

[Br⁻] = 0.3668 M

We need to determine the new concentration of NH_3 again ; So using M_1V_1 =M_2V_2 ; we have :

0.2583 × 750 = M_2 × 810

$M_2=\frac{193.725}{810}\\ M_2 = 0.2392$

$pOH = pKb + log (\frac{HB^+}{HB} )$

$pOH = 4.75 +log(\frac{0.3668}{0.2392})\\pOH=4.75+(0.1857)\\pOH=4.9357$

pH = 14 - pOH

pH = 14 - 4.9357

pH = 9.0643

pH = 9.06

c) After 2.00 M of HBr and Volume = 12.0 mL

Volume increases from 750mL to : (750 + 120 ) mL = 870 mL = 0.87 mL

Definitely the concentration of the Br⁻ will also increase.

The new concentration of [Br⁻] = $\frac{number of moles}{volume}$

moles of Br⁻ added = The new concentration of [Br⁻] × volume

= 120 ×2.0

= 240 mL. mol

mole of Br⁻ already in the solution = 0.1771 + $\frac{240}{1000}$

= 0.1771 + 0.24

= 0.4171 mol of Br⁻

[Br⁻] = $\frac{number of moles}{volume}$

[Br⁻] = $\frac{0.4171}{0.87}$

[Br⁻] = 0.480 M

We need to determine the new concentration of NH_3 again ; So using M_1V_1 =M_3V_3 ; we have :

0.2583 × 750 = M_2 × 870

$M_2=\frac{193.725}{870}\\ M_2 = 0.2227$

$pOH = pKb + log (\frac{HB^+}{HB} )$

$pOH = 4.75 +log(\frac{0.4171}{0.2227})\\pOH=4.75+(0.2725)\\pOH=5.02$

pH = 14 - pOH

pH = 14 - 5.02

pH = 8.98

Answer from: Quest

i think its 100.

hope this !

Answer from: Quest

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;

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A buffer is made by combining 5.55 g of NH3(Kb= 1.8 X 10-5)with 4.78 g of HCl and diluting to a volu...

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