1. Calculate the value of [OH] from the given [H3O+] in each solution and label
ID: 497146 • Letter: 1
Question
1. Calculate the value of [OH] from the given [H3O+] in each solution and label the solution as acidic or basic:
(a) [H3O+] = 4.0 × 10^11 M (b) [H3O+] = 3.3 × 10^3 M.
(a) ____ x10 ____ M, acidic or basic
(b) ____ x10 ____ M, acidic or basic
2. Calculate the value of [OH] from the given [H3O+] in each solution and label the solution as acidic or basic:
(a) [OH] = 3.8 × 10^10 M (b) [OH] = 7.3 × 10^3 M.
(a) ____ x10 ____ M, acidic or basic
(b) ____ x10 ____ M, acidic or basic
3. Calculate the value of [OH] from the given [H3O+] and label the solution as acidic or basic.
(a) 7.8 × 10^2 M (b) 1.8 × 10^9 M
(a) ____ x10 ____ M, acidic or basic
(b) ____ x10 ____ M, acidic or basic
4. What H3O+ concentration corresponds to each pH value: (a) 10.200 (b) 8.700 (c) 3.600?
(a) ____ x10 ____ M
(b) ____ x10 ____ M
(c) ____ x10 ____ M
5. Convert each H3O+ concentration to a pH value.
(a) 4.6 × 10^4 M (b) 6.24 × 10^8 M
Explanation / Answer
Answer :
Ionic Product of Water = [H3O+][HO-] = 1 x 10-14.
Criteria for nature of solution :
If, [H3O+] > [HO-] then Solution is Acidic.
If, [H3O+] = [HO-] then Solution is Neutral..
If [H3O+] < [HO-] then Solution is Basic.
1)
a) [H3O+] = 4.0 x 10-11 M
By eq.(1),
[HO-] = 1.0 x 10-14 / [H3O+]
= 1.00 x 10-14 / 4.0 x 10-11
[HO-]= 2.5 x 10-4 M.
[H3O+] < [HO-] then Solution is Basic.
b) [H3O+] = 3.3 x 10-3 M
By eq.(1),
[HO-] = 1.0 x 10-14 / [H3O+]
= 1.00 x 10-14 / 3.3 x 10-3
[HO-]= 3.03 x 10-12 M.
[H3O+] > [HO-] then Solution is Acidic.
======================================================
2)
a) [HO-] = 3.8 x 10-10 M
By eq.(1),
[H3O+] = 1.0 x 10-14 / [HO-]
= 1.00 x 10-14 / 3.8 x 10-10
[H3O+]= 2.63 x 10-5 M.
[H3O+] > [HO-] then Solution is Acidic.
b) [HO-] = 7.3 x 10-3 M
By eq.(1),
[H3O+] = 1.0 x 10-14 / [HO-]
= 1.00 x 10-14 / 7.3 x 10-3
[H3O+]= 1.37 x 10-12 M.
[H3O+] < [HO-] then Solution is Basic.
======================================================
3)
a) [H3O+] = 7.8 x 10-2 M
By eq.(1),
[HO-] = 1.0 x 10-14 / [H3O+]
= 1.00 x 10-14 / 7.8 x 10-2
[HO-]= 1.28 x 10-13 M.
[H3O+] > [HO-] then Solution is Acidic.
b) [H3O+] = 1.8 x 10-9 M
By eq.(1),
[HO-] = 1.0 x 10-14 / [H3O+]
= 1.00 x 10-14 / 1.8 x 10-9
[HO-]= 5.56 x 10-6 M.
[H3O+] < [HO-] then Solution is Basic.
======================================================
4)
pH = -log[H3O+]
[H3O+] = 10-pH.
a) pH = 10.200
[H3O+] = 10-pH.= 10-10.200 = 6.31 x 10-11.
[H3O+] = 6.31 x 10-11 M
b)
pH = 8.700
[H3O+] = 10-pH.= 10-8.700 = 2.00 x 10-9.
[H3O+] = 2.00 x 10-9 M
c)
pH = 3.600
[H3O+] = 10-pH.= 10-3.600 = 2.51 x 10-4.
[H3O+] = 2.51 x 10-4 M
==============================================
5)
pH = -log[H3O+]
a) [H3O+] = 4.6 x 10-4 M.
pH = -log(4.6 x 10-4)
pH = 3.34
b) [H3O+] = 6.24 x 10-8 M.
pH = -log(6.24 x 10-8)
pH = 7.20
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