Answer:
2
Step-by-step explanation:
A. Moles before mixing
<em>Beaker I:
</em>
Moles of H⁺ = 0.100 L × 0.03 mol/1 L
= 3 × 10⁻³ mol
<em>Beaker II:
</em>
Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.
H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)
[OH⁻] = 0.01 mol·L⁻¹
Moles of OH⁻ = 0.100 L × 0.01 mol/1 L
= 1 × 10⁻³ mol
B. Moles after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 3 × 10⁻³ 1 × 10⁻³
C/mol: -1 × 10⁻³ -1 × 10⁻³
E/mol: 2 × 10⁻³ 0
You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.
You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.
C. pH
[H⁺] = (2 × 10⁻³ mol)/(0.200 L)
= 1 × 10⁻² mol·L⁻¹
pH = -log[H⁺
]
= -log(1 × 10⁻²)
= 2
Well not to sure if that statements true... but in hydrogen for example both its proton and electron have an electric charge but one has a + charge (proton) and the other a - charge (electron) so they end up cancelling so there is no net electrical charge
hope that helps
Answer:
A) The reaction mixture contains mostly products at equilibrium
Explanation:
For the general reaction
Reactants ⇌ Products
The equilibrium constant is
![K_{\text{eq}} = \dfrac{\text{[Products]}}{\text{[Reactants]}}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Beq%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BProducts%5D%7D%7D%7B%5Ctext%7B%5BReactants%5D%7D%7D)
Thus, if K is large, the concentration of products is greater than that of reactants.
The reaction mixture will contain mostly products at equilibrium.
C) and E) are wrong. The equilibrium constant gives us information only on the position of equilibrium, not on how fast it is achieved.
