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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
Once you identify the compound as Ionic<span>, </span>Molecular, or an Acid, follow the individual ... chemicalformulas<span>, write </span>whether<span> the compound is </span>ionic or molecular<span>, and ...</span>
The balanced neutralization reaction here is:
Ca(OH)2 + 2HBr --> 2H2O + CaBr2
Notice that two moles of Her are required to neutralize every one mole of Ca(OH)2. This means that for however many moles of Her reacted, HALF as many moles of Ca(OH)2 reacted as well.
Moles of HBr reacted = 0.75 M x 0.345 L = 0.259 mol
Moles of Ca(OH)2 reacted = 0.259 mol / 2 = 0.130 mol
Concentration of Ca(OH)2 = 0.130 mol / 0.250 L = 0.52 M
Iodine- in most cleaning solutions
