Answer:
6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5
Explanation:
For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).
Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.
HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴
Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).
For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)
= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.
Check using the Henderson - Hasselbalch Equation...
pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18
Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.
pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.
One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.
Answer:
Explanation:
Hello,
At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:
Once we've got the moles we compute the final volume via:
Thus, the molarity of the sodium atoms turn out into:
Now, we perform the same procedure but now for the bromide ions:
Finally, its molarity results:
Best regards.
A: 6 moles
You have to find the molar mass and then multiply
