Answer:
HC₂H₃O₂/KC₂H₃O₂
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pH of the basic buffer solution as:
![ pH=pK_b+log\frac{[salt]}{[acid]} ](https://tex.z-dn.net/?f=%0ApH%3DpK_b%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%0A)
For a best pair, the pKa value must be equal to pH.
NH₃/NH₄Cl forms a basic buffer and cannot account for pH = 5
out of the acidic buffer given,
So, HF , Ka = 3.5 × 10⁻⁴ , So pKa = 3.46
HC₂H₃O₂ , Ka = 1.8 × 10⁻⁵ , So pKa = 4.77
<u>The best pair to show pH = 5 is HC₂H₃O₂/KC₂H₃O₂</u>
Answer:
The answer is 1.15m.
Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.
We can find the number of H2SO4 moles by using its molarity
C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288
Since water has a density of 1.00kgL, the mass of solvent is
m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg
Therefore, molality is
m=nmass.solvent=0.288moles0.250kg=1.15m
