Answer:
pH = 6.999
The solution is acidic.
Explanation:
HBr is a strong acid, a very strong one.
In water, this acid is totally dissociated.
HBr + H₂O → H₃O⁺ + Br⁻
We can think pH, as - log 7.75×10⁻¹² but this is 11.1
acid pH can't never be higher than 7.
We apply the charge balance:
[H⁺] = [Br⁻] + [OH⁻]
All the protons come from the bromide and the OH⁻ that come from water.
We can also think [OH⁻] = Kw / [H⁺] so:
[H⁺] = [Br⁻] + Kw / [H⁺]
Now, our unknown is [H⁺]
[H⁺] = 7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]
[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) / [H⁺]
This is quadratic equation: [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴
a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴
(-b +- √(b² - 4ac) / (2a)
[H⁺] = 1.000038751×10⁻⁷
- log [H⁺] = pH → 6.999
A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.
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Answer:
The answer is given below.
Explanation:
We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.
AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M
HA → H+ + A-
Initial concentration: 0.1 → 3.98 ^10-3 + 0
equilibrium concentration: 0.1(1-α) → 3.98 * 10-3 + 0.1α 0.1α
pKa of chloroacetic acid is 2.9
-log(Ka) = 2.9
Ka = 1.26 * 10-3
From the equation, Ka = [H+] * [A-] / [HA]
1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)
Since α<<1, we assume 1-α = 1
Solving the equation, we have: α = 0.094
Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%
<h3><u>Answer;</u></h3>
0.5 M HBr, pOH = 13.5 ; Has the lowest pH
<h3><u>Explanation;</u></h3>
From the question;
pH = -Log [OH]
or pH = 14 - pOH
Therefore;
For 0.5 M HBr
[H+] = 0.5 M
pH = - Log [0.5]
= 0.30
For; pOH = 13.5
pH = 14 - pOH
= 14 -13.5
= 0.5
For; 0.05 M HCl
pH = - log [H+]
[H+] = 0.05
pH = - Log [0.05]
= 1.30
For; pOH = 12.7
pH = 14 -pOH
= 14 -12.7
= 1.30
For; 0.005 M KOH,
pOH = - log [OH]
[OH-] = 0.005
pOH = - Log 0.005
= 2.30
pH = 14 - 2.30
= 11.7
For; pOH = 2.3
pH = 14 -pOH
= 14- 2.3
= 11.7
