Question

asked Mar 22, 2020 77.3k views

1 Answer

Sloik · Answer 1 · 2020-03-27T09:55:04+0000

Answer: The pH of the solution is 9.14

Step-by-step explanation:

To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 735 torr

V = Volume of the gas = 7.90 L

T = Temperature of the gas =
22^oC=[22+273]K=295K

R = Gas constant =
$62.364\text{ L. torr }mol^(-1)K^(-1)$

n = number of moles of ammonia = ?

Putting values in above equation, we get:

$735torr* 7.90L=n* 62.364\text{ L. torr }mol^(-1)K^(-1)* 295K\\\\n=(735* 7.90)/(62.364* 295)=0.316mol$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of hydrochloric acid = 0.400 M

Volume of solution = 0.450 L

Putting values in above equation, we get:

$0.400M=\frac{\text{Moles of hydrochloric acid}}{0.450L}\\\\\text{Moles of hydrochloric acid}=(0.400* 0.450)=0.18mol$

The chemical reaction for ethylamine and HCl follows the equation:

$NH_3+HCl\rightarrow NH_4Cl$

Initial: 0.316 0.18

Final: 0.136 - 0.18

Volume of the solution = 0.450 L

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(([salt])/([base]))$

$pOH=pK_b+\log(([NH_4Cl])/([NH_3]))$

We are given:

pK_b = negative logarithm of base dissociation constant of ammonia =
$-\log(1.8* 10^(-5))=4.74$

[NH_4Cl]=(0.18)/(0.450)

[NH_3]=(0.136)/(0.450)

pOH = ?

Putting values in above equation, we get:

$pOH=4.74+\log((0.18/0.450)/(0.136/0.450))\\\\pOH=4.86$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-4.86=9.14$

Hence, the pH of the solution is 9.14

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