Answer:
pH =1 2.84
Explanation:
First we have to start with the <u>reaction</u> between HCl and KOH:
Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

We know that
and we have the concentration of the HCl
, when we plug the values into the equation we got:



We can do the same for the KOH values (
and
).



So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Now we can calculate the concentration:


Now, we can <u>calculate the pOH</u> (to calculate the pH), so:
Now we can <u>calculate the pH value</u>: