What is the difference between an acid and its conjugate base?
2 answers:
Answer : The correct option is B: The acid donates an H + ion.
Explanation :
According to Bronsted -Lowry theory, an acid is a substance that donates a proton. The substance that is left after the proton is donated is known as "conjugate base" of that acid.
Consider a general acid HA. The dissociation reaction of the acid can be written as follows.
In the above example, HA behaves as an acid because it donates proton to H₂O. When the proton is donated, A⁻is formed on the product side which is a conjugate base of HA.
From above discussion, we can say that a substance and its conjugate differ by a proton, H⁺ only.
Therefore the acid donates an H + ion is the correct option.
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Answer:
Q(total) = 283Kj
Explanation:
5 Heat Transitions …
Specific Heats => c(s) = 0.50cal/g∙⁰C, c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C
Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g
Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.
Q = mcΔT for temperature change regions of the heating curve (single phase only)
Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)
Solid (ice) => Melting Pt => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal
Melting (s/l) => Liquid (water) => Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal
Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal
Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal
Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal
Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)
= 478cal +7240cal + 9050 cal + 48,870cal + 2036cal
= 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj
Answer:
a) 1 litre of 10% solution and 7 litre of 20% solution
b) 1.67 litres of 50% solution and 8.33 litres of the 20% solution
c) 3.75 litres of 50% solution and 6.25 litres of 20% solution
Explanation:
Given:
chemist needs = 10 liters of a 25% acid solution
Concentration of three solutions that are to be mixed = 10%, 20% and 50%.
Solution:
A) Use 2 liters of the 50% solution
Let us mix this with 10% and 20% solution
They will have to equal 8 litres
Let x=20% solution
Then (8-x) =10%
So the equation becomes,
10%(8-x)+ 20%x+50%(2)=25(10)
(0.1)(8-x) +0.2x+0.50(2)= 0.25(10)
0.8-0.1x+0.2x+1.0=2.5
0.2x-0.1x=2.5-0.8-1.0
0.1x=0.7
x= 7
so, 8-x = 8 -7= 1 litre of 10% solution and 7 litre of 20% solution
B)Use as little as possible of the 50% solution
Let x be the amount of 50% solution.
Then(10-x) be the 20% solution
Now the equation becomes,
50%(x)+20%(10-x)=25%(10)
0.50x+0.2(10-x)=0.25(10)
0.5x+2.0-0.2x=2.5
0.3x=2.5-2.0
0.3x=0.5
x=1.67
now (10-x)=(10-1.67)=8.33
so there will be 1.67 litres of 50% solution and 8.33 litres of the 20% solution
c) ) Use as much as possible of the 50% solution
Let x be the amount of 50% solution.
Then(10-x) be the 20% solution
Now the equation becomes,
50%(x)+10%(10-x)=25%(10)
0.50x+0.1(10-x)=0.25(10)
0.5x+1.0-0.1x=2.5
0.4x=2.5-1.0
0.4x=1.5
x=3.75
Now, (10-x)=(10- 3.75)=6.26
So there will be 3.75 litres of 50% solution and 6.25 litres of 20% solution