According to the following equation, how many ml of 0.15 M NaOH would be needed to titrate 10.00 ml of 0.500 M HCl? 16. What is the pH of a 0.33 M solution of HCI?

Question

asked Jun 1, 2020 204k views

1 Answer

Arielle Nguyen · Answer 1 · 2020-06-06T18:58:10+0000

Answer:

The volume of NaOH required will be 33.33 mL.

0.48 is the pH of a 0.33 M solution of HCI.

Step-by-step explanation:

1)
n_1M_1V_1=n_2M_2V_2

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is
NaOH

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is HCl.

We are given:

$n_1=1\\M_1=0.15 MM\\V_1=?\\n_2=1\\M_2=0.500M\\V_2=10.00mL$

Putting values in above equation, we get:

$1* 0.15 M* V_1=1* 0.500 M* 10.00 mL\\\\V_1=033.33 mL$

The volume of NaOH required will be 33.33 mL.

2) The pH of the solution is defined as negative logarithm of hydrogen ion's concentration in a solution.

$pH=-\log[H^+]$

$pH=-\log[0.33 M]=0.48$

0.48 is the pH of a 0.33 M solution of HCI.