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
pH is defined as the scale on which the substances are classified as acidic or basic. The scale measures from 0 to 14, in which 7 is neutral.
The correct answers are:
From the formula of pH, we have:
pH = -log [H+]
1. To calculate the pH of HCl:
pH = -log [H+ ]
= 1.0 x [tex]10^{-5}[/tex] M HCl
pH = - log [1.0 x [tex]10^{-5}[/tex] ]
pH = 5 - log 1 = 5
2.To calculate the pH of 0.1 M HNO₃, we have:
pH = - log [1.0 x [tex]10^{-1}[/tex] ]
pH = 1 - log 1 = 1
pH = 1
3.To calculate the pH of NaOH, we have:
pH = 1.0 x [tex]10^{-5}[/tex] NaOH
pOH = - log [1.0 x [tex]10^{-5}[/tex] ]
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. To calculate the pH of 0.01 M KOH
pOH = - log ( 1.0 x [tex]10^{-2}[/tex] )
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
Thus, pH is used to determine the acidity and basicity of the compound.
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https://brainly.com/question/15289741
