Calculate the pH of each solution at 25∘C
1 answer:
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
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Explanation:
Given:
20.5 torr
78.6 torr
225 torr
Change into atm
Computation:
We know that;
1 torr = 0.00131579 atm
So,
1. 20.5 torr
= 20.5 x 0.00131579 atm
= 0.02697
= 0.027 atm (Approx)
2. 78.6 torr
= 78.6 x 0.00131579 atm
= 0.103090
= 0.103 atm (Approx)
3. 225 torr
= 225 x 0.00131579 atm
= 0.2960
= 0.296 atm (Approx)
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