The mass of ammonium chloride that must be added is : ( A ) 4.7 g
<u>Given data :</u>
Volume of water ( V ) = 250 mL = 0.25 L
pH of solution = 4.85
Kb = 1.8 * 10⁻⁵
Kw = 10⁻¹⁴
Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :
NH₄CI + H₂O ⇄ NH₃ + H₃O⁺
where conc of H₃O⁺
[ H₃O⁺ ] = 
and Ka = Kw / Kb
∴ Ka = 5.56 * 10⁻¹⁰
Next step : Determine the concentration of H₃O⁺ in the solution
pH = - log [ H₃O⁺ ] = 4.85
∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵
Next step : Determine the concentration of NH₄CI in the solution
C = [ H₃O⁺ ]² / Ka
= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰
= 0.359 mol / L
Determine the number of moles of NH₄CI in the solution
n = C . V
= 0.359 mol / L * 0.25 L = 0.08979 mole
Final step : determine the mass of ammonium chloride that must be added to 250 mL
mass = n * molar mass
= 0.08979 * 53.5 g/mol
= 4.80 g ≈ 4.7 grams
Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g
Learn more about ammonium chloride : brainly.com/question/13050932
Answer:
The boiling point increases with increased pressure up to the critical point, where the gas and liquid properties become identical.
D & e represent the same element
Answer:
1.84 atm
Explanation:
P2=V1P1/V2
Temp is irrelevant because it is constant, so all you have to do is rearrange the ideal gas law.
You can check this by knowing P and V at constant T have an inverse relationship and therefore makes this the correct answer.
- Hope that helps! Please let me know if you need further explanation.
Answer:
The answer to your question is given below.
Explanation:
To prepare 50mL of 3M HCl, we must calculate the volume of the stock solution needed. This can obtained as follow:
Molarity of stock solution (M1) = 12M
Volume of stock solution needed (V1) =?
Molarity of diluted solution (M2) = 3M
Volume of diluted solution (V2) = 50mL
The volume of the stock solution needed can be obtained by using the dilution formula as shown below:
M1V1 = M2V2
12 x V1 = 3 x 50
Divide both side by 12
V1 = (3 x 50)/12
V1 = 12.5mL
The volume of the stock solution needed is 12.5mL
Therefore, to prepare 50mL of 3M HCl, we must measure 12.5mL of the stock solution i.e 12M HCl and then, add water to the mark in a 1L volumetric flask. Now we can measure out 50mL of the solution.
