Answer:
The volume of NaOH required will be 33.33 mL.
0.48 is the pH of a 0.33 M solution of HCI.
Explanation:
1) [tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]NaOH[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is HCl.
We are given:
[tex]n_1=1\\M_1=0.15 MM\\V_1=?\\n_2=1\\M_2=0.500M\\V_2=10.00mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.15 M\times V_1=1\times 0.500 M\times 10.00 mL\\\\V_1=033.33 mL[/tex]
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.
[tex]pH=-\log[H^+][/tex]
[tex]pH=-\log[0.33 M]=0.48[/tex]
0.48 is the pH of a 0.33 M solution of HCI.
