First recognize that NaF is the salt of a strong base (NaOH) and a weak acid (HF), so the pH will be alkaline. Next, look at the hydrolysis of NaF: NaF + H2O ---> NaOH + HF, or looking at it another way.... F^- + H2O ---> HF + OH- and here F^- acts as a base, so we need the Kb for NaF and that will be the inverse of the Ka for HF. The Ka for HF is 6.6x10^-4, so Kb = 1x10^-14/6.6x10^-4 = 1.5x10^-11.
Now, Kb = [HF][OH-]/[F-] = (x)(x)/(0.89) = 1.5x10^-11
x^2 = 1.3x10^-11
x = 3.6x10^-6 = [OH-]
pOH = -log 3.6x10^-6 = 5.44
pH = 8.6 (note the pH is alkaline, as expected)
That's correct. The molar solubility of Na3PO4, NaF, KNO3, AlCl3, and MnS is not affected by the pH of the solution because these compounds do not contain any acidic or basic groups that can significantly influence their solubility as pH changes. The solubility of these compounds is primarily determined by their intrinsic properties and the interactions between the ions in the compound.
When a strong acid is added to a buffer solution containing NaF and HF, the strong acid will react with the weak base (F-) to form HF. The buffer solution will resist changes in pH by the common ion effect, maintaining the solution's acidity around the initial pH of the buffer. The chemical equation can be written as H+ + F- ↔ HF.
- log(1 X 10^-5 M) = 5 14 - 5 = 9 pH ----------
No, pH 2.77 is not the correct pH for 1 M HCl. The pH of 1 M HCl should be 0 (zero) because pH is the negative log the the H+ and for 1 M HCl the [H+] is 1 M, and the negative log of 1 is 0.
pH = -log [H+], so if the [H+] is 2.310 M, the pH = -0.3636
c. The addition of NaF to an aqueous HF solution will increase the concentration of HF. This is because NaF will react with HF to form NaHF2, which increases the amount of HF present in the solution.
To determine the number of moles of NaF in 34.2 grams of a 45.5% by mass solution, first calculate the mass of NaF in the solution. Mass of NaF = 45.5% of 34.2 grams. Then convert the mass of NaF to moles using the molar mass of NaF. Finally, divide the mass of NaF by its molar mass to get the number of moles.
That's correct. The molar solubility of Na3PO4, NaF, KNO3, AlCl3, and MnS is not affected by the pH of the solution because these compounds do not contain any acidic or basic groups that can significantly influence their solubility as pH changes. The solubility of these compounds is primarily determined by their intrinsic properties and the interactions between the ions in the compound.
When a strong acid is added to a buffer solution containing NaF and HF, the strong acid will react with the weak base (F-) to form HF. The buffer solution will resist changes in pH by the common ion effect, maintaining the solution's acidity around the initial pH of the buffer. The chemical equation can be written as H+ + F- ↔ HF.
Naf is not in the Oxford English dictionary.
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
its PH is 3
The conjugate acid of NaF is HF (hydrofluoric acid). When NaF accepts a proton, it forms HF.
Naf-'t'-lee or Naf-'t'-lie
The scientific name for NaF is sodium fluoride.
Britton-Robinson buffer is a "universal" pH buffer used for the range pH 2 to pH 12. Universal buffers consist of mixtures of acids of diminishing strength (increasing pKa) so that the change in pH is approximately proportional to the amount of alkali added. It consists of a mixture of 0.04 M H3BO3, 0.04 M H3PO4 and 0.04 M CH3COOH that has been titrated to the desired pH with 0.2 M NaOH. Britten and Robinson also proposed a second formulation that gave an essentially linear pH response to added alkali from pH 2.5 to pH 9.2 (and buffers to pH 12). This mixture consists of 0.0286 M citric acid, 0.0286 M KH2PO4, 0.0286 M H3BO3, 0.0286 M veronal and 0.0286 M HCl titrated with 0.2 M NaOH.
The pH of a 0.1 M solution of HBr (hydrobromic acid) is around 1. It is a strong acid that dissociates completely in water to produce H+ ions, resulting in a low pH.