what is the pH of 0.15 M NaF?
To determine the pH of a solution, we need to consider the dissociation of the solute. In the case of NaF, sodium fluoride, it is a strong electrolyte and fully dissociates into its constituent ions in aqueous solution.
The dissociation of NaF can be represented as follows:
NaF ↔ Na+ + F-
Since the fluoride ion (F-) is the conjugate base of a weak acid (HF), it can hydrolyze to some extent, resulting in the formation of hydroxide ions (OH-) and hydrogen fluoride (HF) via the following reaction:
F- + H2O ↔ OH- + HF
The hydrolysis of fluoride ions leads to an increase in hydroxide ion concentration (OH-) and, consequently, an increase in the pH of the solution.
To calculate the pH of the solution, we need to determine the hydroxide ion concentration (OH-) resulting from the hydrolysis of the fluoride ion.
The concentration of hydroxide ions is equal to the concentration of fluoride ions that are hydrolyzed. Since NaF fully dissociates into its constituent ions, the initial concentration of fluoride ions (F-) is equal to the concentration of NaF given. Therefore, the initial concentration of hydroxide ions (OH-) can be assumed to be equal to the initial concentration of fluoride ions.
In this case, the concentration of NaF is given as 0.15 M. So, the initial concentration of hydroxide ions is also 0.15 M.
Next, we can use the equation for the hydrolysis of the fluoride ion to determine the concentration of hydrogen fluoride (HF). However, to simplify the calculation, we can assume that the molar concentration of hydroxide ions formed is negligible compared to the initial concentration of fluoride ions. This assumption is reasonable because sodium fluoride is a salt of a strong base (NaOH) and a weak acid (HF), and the hydrolysis of the fluoride ion is only partial.
Therefore, assuming the concentration of hydroxide ions formed is negligible, the concentration of hydrogen fluoride can be taken as equal to the concentration of hydroxide ions formed, which is also equal to the initial concentration of fluoride ions.
Thus, the concentration of hydrogen fluoride in this case is 0.15 M.
At this point, we can enter the concentration of hydrogen fluoride into the equation for the acid dissociation constant (Ka) of HF to calculate the concentration of hydrogen ions (H+).
The Ka for HF is 6.8 x 10^-4. The equation is:
Ka = [H+][F-] / [HF]
Since [H+] is unknown and [F-] = [HF] = 0.15 M, we can rearrange the equation to solve for [H+].
[H+] = Ka * [HF] / [F-]
= (6.8 x 10^-4) * (0.15) / (0.15)
Calculating this, we find that [H+] ≈ 6.8 x 10^-4 M.
The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration. Therefore, we can calculate the pH as follows:
pH = -log[H+]
= -log(6.8 x 10^-4)
≈ 3.17
Hence, the pH of a 0.15 M NaF solution is approximately 3.17.
To find the pH of a solution, we will need to determine whether the given compound is acidic, basic, or neutral.
In this case, NaF refers to sodium fluoride. Sodium fluoride is an ionic compound that dissociates in water to form the sodium cation (Na+) and fluoride anion (F-).
To determine whether the resulting solution is acidic, basic, or neutral, we need to consider the behavior of the ions in water.
The sodium ion (Na+) is the conjugate acid of a strong base (NaOH), but it is a weak acid. Therefore, it does not significantly affect the pH of the solution.
The fluoride ion (F-) is the conjugate base of a weak acid (HF), and it can react with water to produce hydroxide ions (OH-). This reaction is called hydrolysis.
The hydrolysis of fluoride ions (F-) goes as follows:
F- + H2O ↔ HF + OH-
Since fluoride ions (F-) are present in the solution, hydrolysis can occur, producing hydroxide ions (OH-). This hydroxide ion concentration will affect the pH of the solution.
To calculate the pH, we need to determine the hydroxide ion concentration resulting from the hydrolysis. Since we have a 0.15 M NaF solution, we can assume complete dissociation of NaF, and there will be an equal concentration of fluoride ions (F-) and sodium ions (Na+).
However, since the hydroxide ion concentration resulting from the hydrolysis is relatively low, we can use the simplification that the concentration of hydroxide ions is equal to the concentration of fluoride ions.
Therefore, the concentration of hydroxide ions (OH-) produced due to the hydrolysis of the fluoride ions in a 0.15 M NaF solution will also be 0.15 M.
Now, to find the pOH (negative logarithm of the hydroxide ion concentration):
pOH = -log(OH- concentration)
pOH = -log(0.15) = 0.82 (rounded to two decimal places)
Since pH + pOH = 14, we can find the pH:
pH = 14 - pOH = 14 - 0.82 = 13.18 (rounded to two decimal places)
Therefore, the pH of a 0.15 M NaF solution is approximately 13.18.
F^- + HOH ==> HF + OH^-
Kb = (Kw/Ka) = (HF)(OH^-)/(F^-)
Set up an ICE chart, substitute into the Kb expression and solve for OH, convert to pOH, then to pH.