Calculate the pH of the solution formed by mixing 400 mL of 1 M HCl with 100 mL of 4.5 M NaOH.
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Explanation:
Solution:- (B) 12
As we know that,
Molarity of a solution =
volume of solution
(in L)
no. of moles of solute
Given:-
Molarity of HCl solution =0.1M
Volume of HCl solution =40mL=0.04L
Therefore,
No. of moles of HCl=0.04×0.1=0.004 mol
Again,
Molarity of NaOH solution =0.45M
Volume of NaOH solution =10mL=0.01L
Therefore,
No. of moles of NaOH=0.01×0.45=0.0045 mol
Now, for the reaction-
NaOH+HCl⟶NaCl+H
2
O
NaOH is in excess.
Therefore,
Excess amount of NaOH=0.0045−0.004=0.0005 mol
Total volume =0.04+0.01=0.05L
Now,
[OH
−
]=
0.05
0.0005
=0.01M=10
−2
M
Therefore,
pOH=−log[OH
−
]
⇒pOH=−log(10
−2
)=2
Now as we know that,
pH+pOH=14
Therefore,
pH=14−pOH=14−2=12
Hence the pH of the solution is
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