What is the pH?
Calculate the pH of 0.167 M NaC2H3O2 (sodium acetate)
C2H3O2-(aq) + H2O(l) ↔ HC2H3O2(aq) + OH-(aq) Ka for HC2H3O2= 1.75 x 10-5
.............C2H3O2-(aq) + H2O(l) ↔ HC2H3O2(aq) + OH-(aq)
I..............0.167 M................................0...........................0
C.................-x......................................x............................x
E............0.167-x...................................x............................x
Kb for C2H3O2^- = Kw/Ka for HC2H3O2 = (HC2H3O2)(OH^-)/(C2H3O2^-)
Kw/Ka = (x)(x)/(0.167-x). Plug in 1E-14 for Kw and 1.75E-5 for HC2H3O2 and solve for x = (OH^-). Then convert (OH^-) to pH.
Post your work if you get stuck.
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To calculate the pH of a solution, you need to determine the concentration of H+ ions in the solution. In this case, we will use the given information about sodium acetate (NaC2H3O2) and its dissociation reaction.
The dissociation of sodium acetate in water can be represented as follows:
C2H3O2-(aq) + H2O(l) ↔ HC2H3O2(aq) + OH-(aq)
This reaction shows that sodium acetate dissociates into acetate ions (C2H3O2-) and hydroxide ions (OH-) in water.
Given that the concentration of sodium acetate (NaC2H3O2) is 0.167 M, and the dissociation equilibrium constant (Ka) for HC2H3O2 (acetic acid) is 1.75 x 10^-5, we can assume that the reaction has reached equilibrium. This means that the concentration of acetate ions (C2H3O2-) is also 0.167 M.
Now, we can use the equilibrium equation and the equilibrium constant to determine the concentration of hydroxide ions (OH-). In this case, the concentration of hydroxide ions will be the same as the concentration of acetate ions (C2H3O2-).
[H+][OH-] / [HC2H3O2] = Ka
[H+][0.167] / [0.167] = 1.75 x 10^-5
Simplifying the equation, we find:
[H+] = 1.75 x 10^-5 M
The concentration of hydrogen ions ([H+]) is equal to the concentration of hydroxide ions ([OH-]) because the dissociation of sodium acetate produces equal quantities of these ions.
Finally, we can use the equation pH = -log[H+] to calculate the pH value:
pH = -log(1.75 x 10^-5) ≈ 4.76
Therefore, the pH of a 0.167 M solution of sodium acetate is approximately 4.76.