(P1V1/T1)=(P2V2/T2)
I just took the test and this should be correct
Note the signs of equilibrium:-
- Reaction don't procede forward or backward
- Concentration of products and reactants remains same .
So
if
Concentration of A is 2M then concentration of B should be same .
So equilibrium constant K is 1
![\\ \rm\rightarrowtail K=\dfrac{[Products]^a}{[Reactants]^b}](https://tex.z-dn.net/?f=%5C%5C%20%5Crm%5Crightarrowtail%20K%3D%5Cdfrac%7B%5BProducts%5D%5Ea%7D%7B%5BReactants%5D%5Eb%7D)
So
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
