Question

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)?

Answer 1

Answer:

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 2

Buffer is the solution that can neutralize the added acids or base in them. pH is 9.41, after adding 60 ml it is 9.06 and after adding 120 ml it is 8.98.

What is pH?

pH is the concentration of the hydrogen ion in the solution. It gives the acidity and the basicity of the solution.

Given,

Mass of Ammonia = 5.55 g

Mass of hydrochloric acid = 4.78 gm

The molar mass of ammonia = 17 g/mol

Molar mass hydrochloric acid = 36 g/mol

The moles of ammonia will be  $\dfrac{5.55}{17} = 0.3265 \rm\; mol.$

The moles of hydrochloric acid will be $\dfrac{4.78}{36}= 0.1328 \;\rm mol.$

The final concentration of ammonia is  0.3265 - 0.1328 = 0.1937 mole and, final concentration of HCl is 0.1328 -0.1328 = 0.00 mole.

The concentration of ammonia is given as:

$\begin{aligned}\rm Concentration &= \rm \dfrac{moles}{Volume}\\\\&= \dfrac{0.1937}{0.75}\\\\&= 0.2583\;\rm mol/L\end{aligned}$

The concentration of ammonium chloride is given as:

$\begin{aligned}\rm Concentration &= \rm \dfrac{moles}{Volume}\\\\&= \dfrac{0.1328}{0.75}\\\\&= 0.1771\;\rm mol/L\end{aligned}$

The pOH can be calculated by the Henderson Hasslelbach equation as:

$\begin{aligned}\rm pOH &= \rm pKb + log \dfrac{[HB^{+}]}{[HB]}\\\\&= \rm -log(1.8\times 10^{-5}) + log (\dfrac{0.1771}{0.2583})\\\\&= 4.5861\end{aligned}$

pH can be estimated as:

$\begin{aligned}\rm pH &= \rm 14 - pOH\\\\\rm pH &= 14 - 4.5861\\\\\rm pH &= 9.41\end{aligned}$

After the addition of 60 mL of 2.00 M of HBr, pH can be calculated by calculating pOH.

The pOH can be calculated by the Henderson Hasslelbach equation as:

$\begin{aligned}\rm pOH &= \rm pKb + log \dfrac{[HB^{+}]}{[HB]}\\\\&= \rm 4.75 +(0.1857)\\\\&= 4.937\end{aligned}$

pH can be determined as:

$\begin{aligned}\rm pH &= \rm 14 - pOH\\\\\rm pH &= 14 - 4.9357\\\\\rm pH &= 9.06\end{aligned}$

After the addition of 120 mL of 2.00 M of HBr, pH can be calculated by calculating pOH.

The pOH can be calculated by the Henderson Hasslelbach equation as:

$\begin{aligned}\rm pOH &= \rm pKb + log \dfrac{[HB^{+}]}{[HB]}\\\\&= \rm 4.75 +(0.2725)\\\\&= 5.02\end{aligned}$

pH can be determined as:

$\begin{aligned}\rm pH &= \rm 14 - pOH\\\\\rm pH &= 14 - 5.02\\\\\rm pH &= 8.98\end{aligned}$

Therefore, the pH is 9.41, 9.06 and 8.98.

Learn more about pH here:

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