have : [NaF] = 0.22 M
Ka of HF is 3.5 × 10-5
pH =?
from:
NaF(s) dissolve in water give: Na+(aq)+F-(aq)
thus,F-(aq)+H2O(l) <---------->HF(aq)+OH-(aq)
0.22 M00
if0.22 - BBB(mol/L)
Ka = [HF].[OH-]/[F-], but [OH-] is less we give [F-] = 0.22 - B≈ 0.22 M
3.5 x 10-5 = B2/0.22 --> B = [OH-] = (3.5 x 10-5x0.22)1/2≈2.775x10-3 M
we have; pOH = -log[OH-] and pH + pOH = 14
-> pH = 14 -pOH = 14 + log[OH-] = 14 - log(2.775x10-3)≈ 11.443
thus, the pH of a 0.22 M NaF is pH = 11.443
Determine the pH of a 0.22 M NaF solution. The Ka of HF is 3.5 × 10-5
Determine the pH of a solution that is 1.00 L of 0.100 M HF and 0.100 M NaF after 0.50 mole of solid KOH has been added to the solution. Ka(HF) = 3.5×10-4
2. HF ionizes in water to produce H.O'(aq) and F (aq) ions. 0.22 M Naf solution is added to a 0.10 M HF(aq) at 25°C. The Ka of HF is 3.5 x 10-5. (Hint: NaF is a salt. F is the common ion for the equilibrium). Calculate the pH of the solution. Show work.
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Calculate the pH of a 3.27×10-3 M solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.
5) A 1 L buffer solution is 0.520 M in HF and 0.520 M in NaF. Calculate the pH of the solution after adding 0.220 moles of NaOH. Assume no volume change upon the addition of a base. Ka for HF is 3.5 X 10-4
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A 1.0-liter solution contains 0.25 M HF and 0.32 M NaF (Ka for HF is 7.2 × 10–4). What is the pH of this solution
help with 5 and 6 please 5. Choose the effective pH range of a HF/Naf buffer. For HF, Ka = 3.5 x 10-4 6. If a small amount of a strong acid is added to buffer made up of a weak acid, HA, and the sodium salt of its conjugate base, NaA, the pH of the buffer solution does not change appreciably because a) the Ka of HA is changed. b) the strong acid reacts with Ato give HA, which...
A buffer solution contains 1.0 M HF and 1.0 M NaF. The ka for HF is 7.2x10-5. 0.10 moles of HCl are added to 1 liter of the buffer. The pH of the resulting solution is a) 4.05 b) 4.14 c) 4.23 d) 4.74