1)calculate [OH-] and pH for each of the following solutions. (Kw = 1.0*10^-14.)
a)0.101 M NaF
b) 0.030 M Na2S
c)a mixture that is 0.019 M in CH3COONa and 0.036 M in (CH3COO)2Ba
please help and if u can try to explain
To calculate the hydroxide ion concentration ([OH-]) and pH for each of the given solutions, you need to consider the dissociation of the compounds in water and then use the equation relating [OH-] and pH.
a) 0.101 M NaF:
When NaF is dissolved in water, it dissociates into Na+ and F- ions. F- reacts with water partially to form OH- ions and HF (hydrofluoric acid). As HF is a weak acid, it does not dissociate completely. Therefore, we can assume that all of the F- ions react with water.
Since NaF is Na+ and F-, the initial concentration of [F-] is equal to the concentration of NaF, which is 0.101 M. As all of the F- reacts with water, the concentration of OH- formed is also 0.101 M.
To calculate the pH, you can use the equation: pH = -log[H+]. In this case, [H+] is equal to [OH-], which is 0.101 M. Thus, pH = -log(0.101) = 1.995.
b) 0.030 M Na2S:
Na2S dissociates in water to produce two Na+ ions and one S2- ion. The S2- ion reacts completely with water to form OH- ions.
Since the concentration of Na2S is 0.030 M, the concentration of S2- ions is also 0.030 M. Therefore, the concentration of OH- ions formed is also 0.030 M.
Using the equation pH = -log[H+], we know that [H+] is equal to [OH-], which is 0.030 M. Hence, pH = -log(0.030) = 1.522.
c) A mixture of 0.019 M CH3COONa and 0.036 M (CH3COO)2Ba:
When CH3COONa and (CH3COO)2Ba dissociate in water, they produce CH3COO- ions. The CH3COO- ions react partially with water to form OH- ions and acetic acid (weak acid). The initial concentrations of CH3COO- ions in the mixture are 0.019 M and 0.036 M for CH3COONa and (CH3COO)2Ba, respectively.
To calculate the concentration of OH- ions, we need to find the excess OH- ions formed due to the reaction between CH3COO- ions and water. This requires determining the common ion effect.
By comparing the initial concentrations of CH3COO- ions from both compounds and using stoichiometry, the excess concentration can be found. In this case, the concentration of OH- ions formed will be the difference between the initial concentrations of CH3COO- ions from CH3COONa and (CH3COO)2Ba.
Once you have the excess concentration of OH- ions, use the equation pH = -log[H+] to find the pH.
Remember to double-check the thermodynamic dissociation constants for acetic acid, as it will be used to calculate the concentration of OH- ions formed.
I hope this explanation helps you understand how to approach these calculations.
a and b are salts. The pH of those solutions is determined by the hydrolysis of the salt. c is a mixture; you calculate the molarity of the acetate ion and go from there. Here is how you do the first one, in detail.
.............F + HOH ==> HF + OH^-
initial.....0.1..........0......0
change.......-x..........x.......x
equil.......0.1-x.........x......x
Kb for F^- = (Kw/Ka for HF) = (HF)(OH^-)/(HF) = (x)(x)/(0.1-x)
and solve for x = (OH^-) = (HF) and convert to pH.
A shortcut I use for calculating pH and pOH and that stuff is, if x is OH^-, as it will be in all of these calculations, I take pOH = -log(OH^-), then convert to pH by
pH + pOH = pKw = 14. So while the pOH is still in the calculator (as a negative number), and without changing the sign, I key in + 14 and out pops the pH.