Question

What is the Theoretical pH of 0.10 grams of sodium acetate with a volume of 10.0mL?

Answer 1

Well, assuming an aqueous solution of this exists, then the theoretical `pH` would be `12`.

Explanation:

We first need to calculate the molarity of this solution.

Sodium acetate has a chemical formula of `CH_3COONa`, and therefore has a molar mass of around `82g"/"mol`.

Molarity is expressed by the formula

`"molarity"="moles of solute"/"liters of solution"`

We therefore need to convert the sodium acetate into moles, and the volume into liters.

`0.1g` of sodium acetate would be

`(0.1cancelg)/(82cancelg"/"mol)=0.0012195122~~1.22*10^-3 \ mol`

Then, `10mL=0.01L=1*10^-2 \ L`

`:."molarity"=(1.22*10^-3 \ mol)/(1*10^-2 \ L)=0.122 \ mol"/"L=0.122M`

Since sodium acetate is a salt formed by neutralization of acetic acid and sodium hydroxide (see the equation):

`CH_3COOH(aq)+NaOH(aq)->CH_3COONa(aq)+H_2O(l)`

where `CH_3COONa` is sodium acetate.

Acetic acid is a weak acid, while sodium hydroxide is a strong base, so an aqueous sodium acetate solution would be slightly basic.

Therefore, we would need to calculate the `pOH` of the solution.

`pOH=-log[OH^-]` where `[OH^-]` is the molarity of the solution.

So, the `pOH` of this solution would be

`pOH=-log[0.122]~~0.91`

We know that at `25^@C`,

`pH+pOH=14`

`:.pH=14-0.91=12.09~~12`

So, the solution would have a `pH` of `12`.

Answer 2

Well, we solve the equilibrium expression....but we need more data.

`AcO^(-) + H_2O(l) rightleftharpoonsAcOH + HO^-`

Explanation:

And as always for equilibria...

`K_b=([HO^(-)][AcOH])/([AcO^(-)])`

Of course, you have not quoted `K_b`..but `K_b(AcO^(-))=5.75xx10^-10`...

And so `([AcOH][HO^-])/([AcO^(-)])=5.75xx10^-10`

And if we say that `x*mol*L^-1` acetate ion ASSOCIATES then...

`x^2/(((0.10*g)/(82.03*g*mol^-1))/(10xx10^-3*L)-x)=5.75xx10^-10`

We could solve for `x` exactly....but we make the approximation that `((0.10*g)/(82.03*g*mol^-1)-x)/(10xx10^-3*L)~=((0.10*g)/(82.03*g*mol^-1))/(10xx10^-3*L)`

And so `x~=sqrt(5.75xx10^-10xx((0.10*g)/(82.03*g*mol^-1))/(10xx10^-3*L))`

`x_1=8.37xx10^-6*mol*L^-1`

`x_2=8.37xx10^-6*mol*L^-1`

But `x=[HO^-]`...and thus `pOH=-log_10(8.37xx10^-6)=5.08`..and so `pH=14-pOH=14-5.08=8.92`.

`pH` is slightly basic as expected....