Answer:
The relationship is expressed as follows: ![K_{a} = \frac{[H+][A-]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%2B%5D%5BA-%5D%7D%7B%5BHA%5D%7D)
Explanation:
Most acidic substances are weak acids and are therefore only partially ionized in acqeous solution. We cab use the equilibrium constant for the ionization of acid to express the extent to which the weak acid ionizes. If we represent a general weak acid as HA, we can write the equation for its ionization reaction like this:
To calculate the pH of a weak acid, we use the equilibrium concentration of the reacted species and product.
Take for example:
HA → H + A⁻
where A id the conjugate base.
Knowing that x amount of acid reacts, we can solve like this:
HA → H + A⁻
H+ = antilog (pH)
thus, the pH of the acid is equals to H+ (initial) - H+ (equilibrium) ≈ H+ (initial)
Answer:
The dog should be given half of 100 mg tablet in the morning and another half of 100 mg tablet in an evening.
Explanation:
Weight of the dog = 50 pounds = 22.68 kg
1 kg = 2.205 pounds
Amount of daily dose prescribed by doctor = 4.4 mg/kg
Total mount of dose = 4.4 mg/kg × 22.68 kg = 99.79 mg
99.79 mg ≈ 100 mg
49.89 mg ≈ 50 mg
So, according to doctor prescription dog should be given half 100 mg tablet in the morning and another half of 100 mg tablet in an evening.
ZnSO₄ + 3NH₃ = [Zn(NH₃)₃]SO₄ threeamminezinc sulfate
Zn²⁺ + 3NH₃ = [Zn(NH₃)₃]²⁺ threeamminezinc cation
<span>True Solution is a homogeneous mixture of two
or more substances in which substance dissolved (solute) in solvent has
the particle size of less than 10-9 m or 1 nm. Simple solution of sugar in water is an example of true solution.</span>
Answer:
130.4 grams of sucrose, would be needed to dissolve in 500 g of water.
Explanation:
Colligative property of boiling point elevation:
ΔT = Kb . m . i
In this case, i = 1 (sucrose is non electrolytic)
ΔT = Kb . m
0.39°C = 0.512°C/m . m
0.39°C /0.512 m/°C = m
0.762 m (molality means that this moles, are in 1kg of solvent)
If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles
Molar mass sucrose = 342.30 g/m
Molar mass . moles = mass
342.30 g/m . 0.381 m = 130.4 g
