If 16 mL of 0.78 M HCl is added to 109 mL of 0.20 M NaOH, what is the final pH?
A) 0.76
B) 13.24
C) 12.87
D) 7.00
E) 1.13

Question

Quizplus · Accepted Answer

First, we need to find out how many moles of NaOH we have:

0.109 L x 0.20 mol/L = 0.0218 mol NaOH

Next, we need to find out how many moles of HCl we have:

0.016 L x 0.78 mol/L = 0.01248 mol HCl

Since the balanced chemical equation for the reaction between HCl and NaOH is:

HCl + NaOH → NaCl + H2O

we can see that one mole of HCl reacts with one mole of NaOH. Since we have more NaOH than HCl, not all of the NaOH will react and some will be left over.

The amount of NaOH that will react with the HCl is:

0.01248 mol NaOH (since 1 mol HCl reacts with 1 mol NaOH)

So the amount of NaOH that will be left over is:

0.0218 mol NaOH - 0.01248 mol NaOH = 0.00932 mol NaOH

To find the concentration of the excess NaOH, we need to divide the number of moles by the total volume:

0.00932 mol / 0.125 L = 0.0746 M

Now we can use the concentration of the excess NaOH to find the pOH:

pOH = -log(0.0746) = 1.13

Finally, we can use the definition of pH to find the pH:

pH = 14 - pOH = 12.87

Therefore, the final pH is 12.87, which is answer choice C.

If 16 ML of 0