Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
The pH of the solution is 9.14
For ammonia:
To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:
PV = nRT
P= Pressure of the gas = 735 torr
V= volume of the gas = 7.70 l
T= Temperature of the gas= 22°C (22+273K) = 295K
R= Gas constant= 62.364 L torr/mol/K
n = number of moles of ammonia = ?
By substituting the values in the above equation, we get:
735×7.70 = n × 62.364 × 295
5659.5 = n × 18397.38
n= 5659.5/18397.38
n= 0.3076
For hydrochloric acid:
To calculate the number of moles for given molarity, we use the equation:
Molarity of the solution = [tex]\frac{moles of solute}{volume of solution}[/tex]
Molarity of hydrochloric acid = 0.400 M
Volume of solution = 0.250 L
Putting values in above equation, we get:
0.400= moles of solute/0.250
moles of hydrochloric acid = 0.1 mol
The chemical reaction for ethylamine and HCl follows the equation:
NH3+HCl---> NH4Cl
Initial: 0.316 0.18
Final: 0.136 - 0.18
volume of the solution = 0.250 l
To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:
pOH = pKb + log([tex]\frac{salt}{base}[/tex])
pOH = pKb + log([tex]\frac{NH4Cl}{NH3}[/tex])
it is given:
pKb = negative logarithm of base dissociation constant of ammonia =
-log(1.8 × [tex]10^{-5}[/tex]) = 4.74
[NH4Cl] = 0.18/0.450
[NH3] = 0.136/0.450
On substituting the values we get,
pOH= 4.74 + log ([tex]\frac{0.18/0.250}{0.136/0.250}[/tex])
pOH = 4.74 + 0.12
pOH= 4.86
pH + pOH= 14
pH= 14-pOH
= 14-4.86
pH = 9.14
pH of the solution = 9.14
To learn more about calculating pH visit:
https://brainly.com/question/9278932
#SPJ4
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.