00:01
Hi! so to solve for this, we'll write first the ice table for the dissociation of hf and the fluoric acid.
00:13
This will dissociate to h plus and f minus.
00:19
Now initial change equilibrium concentration.
00:23
The initial concentration of the solution is unknown so we'll just call that y and then products are zero initially.
00:31
This will lose some amount, minus x.
00:34
This will gain plus x at the equilibrium.
00:37
We'll have initial minus change so y minus x, x and x.
00:42
And our unknown here is the amount of hf so that means we need to solve for y first before we could solve for the mass of hydrofluoric acid.
00:52
So we could solve for the value of x here using the given ph of the solution.
00:59
Remember that the hydrogen ion concentration is equivalent to 10 raised to the power of negative ph.
01:08
Sorry let me rewrite this one.
01:10
This is a bracket.
01:17
Alright that's better.
01:19
Now we could plug in the value that we have 10 raised to the power of negative 2 .49 and this will give us 3 .24 times 10 to the negative 3 molar.
01:37
Now remember that as per our ice table, h plus is equivalent to x at equilibrium.
01:43
Therefore x is equivalent to 3 .24 times 10 to the negative 3 and same for this one.
01:52
And you can write this as this is equivalent to y minus 3 .24 times 10 to the negative 3.
02:01
And using the given ka to solve for the value of y, remember that ka is equivalent to the concentration of the products.
02:08
So our products we have h plus and f minus over the initial concentration, equilibrium concentration rather of the reactant.
02:21
So these are concentrations at equilibrium.
02:23
Now let's plug in the information that we have...