Question

Titration Problem - It took me 25.4 mL of a 5.0 M H2SO4 to titrate 50.0 mL of a KOH solution with unknown concentration, what was the concentration and the pH of the basic solution?

Answer 1

Concentration is `5.08 \ "M"`, and the `"pH"` of the solution is around `14.7`.

Explanation:

Well, we got the balanced equation:

`H_2SO_4(aq)+2KOH(aq)->K_2SO_4(aq)+H_2O(l)`

So, two moles of potassium hydroxide are needed to liberate one mole of sulfuric acid.

Now, I'll calculate the moles of sulfuric acid.

Using the molarity formula,

`c=n/V`

• `n` is the number of moles

• `V` is the volume, usually in liters

Rearranging,

`n=c*V`

And so,

`n=0.0254color(red)cancelcolor(black)"L"*(5 \ "mol")/(color(red)cancelcolor(black)"L")=0.127 \ "mol"`

Since we need double the amount of base to neutralize the acid, we would need to use `0.127 \ "mol"*2=0.254 \ "mol"` of base.

Also, it tells us that we used `50 \ "mL"=0.05 \ "L"`.

We again can use the molarity equation, but this time, we want the concentration, and we get:

`c=(0.254 \ "mol")/(0.05 \ "L")`

`=5.08 \ "mol/L"`

`=5.08 \ "M"`

Now, to find the `"pH"`, we need to first write down the dissociation of the base.

`KOH(aq)->K^+(aq)+OH^(-)(aq)`

So, `5.08` moles per liter of potassium hydroxide would have `5.08` moles of hydroxide ions.

We first need to find the `"pOH"` of the solution. It's given by:

`"pOH"=-log[OH^-]`

• `[OH^-]` is the concentration of hydroxide ions in terms of molarity

And so, `"pOH"=-log(5.08)`

`~~-0.71`

`"pH"` and `"pOH"` are related by:

`"pH + pOH"=14`

Therefore, `"pH"=14-(-0.71)`

`~~14.7`