Physical change - No change of matter in this phase
chemical change - All types of phase change occur here
Answer:
B
The increase in the atomic number
Answer is: the maximum concentration of Pb²⁺ is 6.8·10⁻³ M.
Chemical reaction 1: PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq).
Chemical reaction 2: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Ksp(PbCl₂) = 1.7·10⁻⁵.
c(NaCl) = c(Cl⁻) = 0.0500 M.
Ksp(PbCl₂) = c(Pb²⁺) · c(Cl⁻)².
c(Pb²⁺) = Ksp(PbCl₂) ÷ c(Cl⁻)².
c(Pb²⁺) = 1.7·10⁻⁵ M³ ÷ (0.0500 M)².
c(Pb²⁺) = 0.000017 M³ ÷ 0.0025 M².
c(Pb²⁺) = 0.0068 M = 6.8·10⁻³ M.
Answer:
pKa = 4.89.
Explanation:
We can solve this problem by using the <em>Henderson-Hasselbach equation</em>, which states:
pH = pKa + log ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.
We <u>input the given data</u>:
4.63 = pKa + log 
And <u>solve for pKa</u>:
pKa = 4.89