Answer:
pH is the negative logarithm of hydrogen (oxonium) ion concentration of aqueous solutions.
Explanation:
Recall that pH is defined as the negative logarithm of hydrogen ion concentration. We are going to apply that definition here.
a) 0.00010 M HClO4
pH= -log[0.00010]
pH= 4
b) 2.35 x 10-3 M HI
pH= -log[2.35 x 10-3 ]
pH= 2.63
c) 4.678 x 10-5 M HNO3
pH= -log[4.678 x 10-5]
pH= 4.3
d) 0.0000877 M HCl
pH=-log[0.0000877]
pH= 4.1
e) 0.00020 M HF
pH= -log[0.00020]
pH= 3.7
f) 0.0000061 M H3O+
pH= -log[0.0000061]
pH= 5.2
g) 4.3 x 10-3 M NaOH
pOH= -log[OH^-]
pOH= -log[ 4.3 x 10-3 ]
pOH= 2.4
But
pH + pOH =14
Therefore
pH= 14-pOH
pH= 14 -2.4
pH= 11.6
h) 6.7×10^-9 M KOH
pOH= -log[OH^-]
pOH= -log[ 6.7 x 10-9 ]
pOH= 8.2
But
pH + pOH =14
Therefore
pH= 14-pOH
pH= 14 -8.2
pH=5.8
I) 0.00044M HCl
pH= -log[0.00044]
pH= 3.4
Answer: A. 4 B. 2.6 C. 4.3 D. 4.1 E. 3.7 F. 5.8 G. 11.6 H. 5.8 I. 3.4
Explanation:
A. 0.00010 M HClO4
pH = -log [1.0 x 10^-4]
= 4 - log 1 = 4
B. 2.5 x 10^-3 M HI
pH = -log [2.5 x 10^-3]
= 3 - log 2.5
= 3 - 0.398 = 2.6
C. 4.678 x 10^M HNO3
pH = -log [4.678 x 10^-5]
= 5 - 0.7 = 4 .3
D. 0.0000877M HCL
pH = - log [8.77 x 10^-5]
= 5 -0.943 = 4.057 =4.1
E. 0.00020M HF
pH = -log [2.0 x10^-4]
= 4 - 0.301 = 3.7
F. 0.0000061M H3O+
pH = - log [ 6.1 x 10^-6]
= 6 -0.785 = 5.2
G. 4.3 x 10^-3 M NaOH
Finding pOH first , we have
pOH = - log [4.3 x 10-3]
= 3 - 0.633 = 2.4
pH + pOH = 14
pH = 14 - 2.4 = 11.6
H. 6.7 x 10^-9 M KOH
Finding pOH, we have,
pOH = - log [6.7 x 10^-9]
= 9 - 0.826 = 8.17 = 8.2
pH + pOH = 14
pH = 14 - 8.2 = 5.8
I. 0.00044 M HCl
pH = -log [ 4.4 x 10^-4]
= 4 - 0.64 = 3.36 = 3.4
