Both D and G options. Weak base and its conjugate acid or weak acid and its conjugate base are the possible components of a buffer solution.
Explanation:
Buffer solution is the solution which gets easily dissolved in water and so called as "Aqueous solution".
Buffer solution is essentially made up of two components Known as:
i.) Weak base and its conjugate acid
ii.) Weak acid and its conjugate base
This weak acid and base solution is used to maintain the pH value of the solution in a balanced way.
When the weak acid or base solution is added to strong acid or base solution that is the way pH gets balanced .
In one word buffer solution is the solution which resists for the pH change when strong acids or bases are added.
Answer:
The pH of the solution is 1.38.
Explanation:
Mass of HCl = 614 mg = 0.614 g
Moles of HCl = 
Concentration of HCl :
On adding 0.01682 moles to 400 mL of water that 0.4 L of water.
![[HCl]=\frac{0.01682 mol}{0.4 L}=0.04205 M](https://tex.z-dn.net/?f=%5BHCl%5D%3D%5Cfrac%7B0.01682%20mol%7D%7B0.4%20L%7D%3D0.04205%20M)
1 mole of HCl gives 1 mole of hydronium ion and 1 mole of chloride ions in an aqueous solution.
Then 0.04205 mol/L of HCl will give:
of hydronium ions.
![[H_3O^+]=0.04205 M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.04205%20M)
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
![pH=-\log [0.04205 M]=1.38](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5B0.04205%20M%5D%3D1.38%20)
The pH of the solution is 1.38.
a) f2 loses electrons and is oxidised. li gains electrons and is reduced
c)sn2+ loses electrons and is oxidised.Al gains electrons and is reduced
b)br2 loses electrons and is oxidised. I- in KI gains electrons and is reduced
Answer:
1.87x10⁻³ M SO₄²⁻
Explanation:
The reaction of SO₄²⁻ with Ba²⁺ (From Ba(NO₃)₂) is:
SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s)
<em>Where 1 mole of SO₄²⁻ reacts per mole of Ba²⁺</em>
<em />
To reach the end point in this titration, we need to add the same moles of Ba²⁺ that the moles that are of SO₄²⁻.
Thus, to find molarity of SO₄²⁻ we need to find first the moles of Ba²⁺ added (That will be the same of SO₄²⁻). And as the volume of the initial sample was 100mL we can find molarity (As ratio of moles of SO₄²⁻ per liter of solution).
<em>Moles Ba²⁺:</em>
7.48mL = 7.48x10⁻³L ₓ (0.0250moles / L) = 1.87x10⁻⁴ moles of Ba²⁺ = Moles of SO₄²⁻
<em>Molarity SO₄²⁻:</em>
As there are 1.87x10⁻⁴ moles of SO₄²⁻ in 100mL = 0.1L, molarity is:
1.87x10⁻⁴ moles of SO₄²⁻ / 0.1L =
<h3> 1.87x10⁻³ M SO₄²⁻</h3>
An ion is an atom that has lost or gained an electron so the relation ship you would see would be a negative or a positive one depending on if it gained or lost an electron.<span />
