calculate the pHof 10 raise to power -8 M HCI solutions
Answers
Step-by-step explanation:
On using this relation, pH= –log[H
3
O
+
], we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7.
It may be noted that in very dilute acidic solution when H
+
concentrations from acid and water are comparable, the concentration of H
+
from water cannot be neglected.
Therefore,
[H
+
] total = [H
+
] acid + [H
+
] water
Since HCl is a strong acid and is completely ionized.
[H
+
]HCl=1.0×10
−8
The concentration of H
+
from ionization is equal to the [OH
–
] from water,
[H
+
]H
2
O=[OH
–
]H
2
O = x (say)
[H
+
] total =1.0×10
−8
+x
But,
[H
+
][OH
–
]=1.0×10
−14
(1.0×10
−8
+x)(x)=1.0×10
−14
x
2
+10
−8
x –10
−14
=0
Solving for x, we get x=9.5×10
−8
Therefore,
[H
+
]=1.0×10
−8
+9.5×10
−8
=10.5×10
−8
=1.05×10
−7
pH= –log[H
+
]
= –log(1.05×10
−7
)=6.98
In the case of very dilute acid like 10^-8 M HCL, the H+ concentration of acid is equal to or less than the H+ concentration of water hence
total [H+]=[H+]acid+[H+]water
=1*10^-8+10^-7
=1.1*10^-7
PH=-log10[H+]
=6.9586
Note: any acid must have ph less than 7