Answer: 4.21×10⁻⁸
Explanation:
1) Assume a general equation for the ionization of the weak acid:
Let HA be the weak acid, then the ionization equation is:
HA ⇄ H⁺ + A⁻
2) Then, the expression for the ionization constant is:
Ka = [H⁺][A⁻] / [HA]
There, [H⁺] = [A⁻], and [HA] = 0.150 M (data given)
3) So, you need to determine [H⁺] which you do from the pH.
By definition, pH = - log [H⁺]
And from the data given pH = 4.1
⇒ 4.10 = - log [H⁺] ⇒ [H⁺] = antilog (- 4.10) = 7.94×10⁻⁵
4) Now you have all the values to calculate the expression for Ka:
ka = 7.94×10⁻⁵ × 7.94×10⁻⁵ / 0.150 = 4.21×10⁻⁸
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Br2 is the correct answer
In order to represent a noble gas the Orbital notation should be equal to 2, 8 or 18 for eg. 1s²
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
