Answer:
The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases
Explanation:
Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in solution decreases significantly.
In this case, both BaSO4 and Na2SO4 both possess the SO4^2- anion. Hence SO4^2- anion is the common ion. Given the equilibrium;
BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq), addition of Na2SO4 will decrease the solubility of BaSO4 due to the presence of a common SO4^2- anion compared to pure water.
This implies that the equilibrium will shift to the left, (more undissoctiated BaSO4) hence decreasing the Ba^2+(aq) concentration.
3 elements are present in HCOOH - hydrogen, oxygen, and carbon.
Grease is nonpolar. It is made up of lengthy hydrocarbon chains attached to glycerol (triacylglycerols). Because of this property, nonpolar solvents, such as tetrachloroethylene, can get grease out of clothes. This solvent acts by dissolving the grease, thereby removing it from clothes without leaving any residue.
Answer:I think it could be A or B but I would choose A.
Explanation:
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
