Question

When 100 ml of N/10 NaOH are added to 50 ml of N/5 HCl, the pH of the resulting solution is:
a.) 7
b.) greater than 7
c.) less than 7
d.) Zero

Answer 1

Hint: From given normality and volume values, we can calculate the equivalents of solution and then by knowing which component acid or base is more in solution, we can find their pH. Acids have pH less than 7 and base has pH more than 7. The neutral solution has pH 7.

Complete answer:
First, we will write what is given to us and what we need to find out.
Given :
Normality of NaOH = $\dfrac{1}{{10}}$ N
Volume of NaOH = 100 ml
Normality of HCl = $\dfrac{1}{5}$ N
Volume of HCl = 50 ml
To find :
pH of the resulting solution
We know Equivalents of NaOH = Normality of NaOH $ \times $Volume of NaOH
Equivalents of NaOH = $\dfrac{1}{{10}}$ $ \times $100
Equivalents of NaOH = 10 milli equivalent
We can find for HCl as -
Equivalents of HCl = Normality of HCl $ \times $Volume of HCl
Equivalents of HCl = $\dfrac{1}{5}$ $ \times $50
Equivalents of HCl = 10 milli equivalent
We observe that
Equivalents of NaOH = Equivalents of HCl
This means that NaOH has neutralised HCl. So, the pH will also be of neutral solution.
We know the pH of neutral solution is 7.
Thus, the pH of the resulting solution is 7.

So, the option a.) is the correct answer.

Note:
It must be noted that HCl is a strong acid and NaOH is a strong base. When both react in equimolar amounts, they will result in a neutral solution. So, the pH would be 7. If the amount of HCl would have been more than the pH would have been less than 7 and if the amount of NaOH would have been more, then the pH would have been more than 7.