PH = pKa + log
![\frac{[base]}{[Acid]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bbase%5D%7D%7B%5BAcid%5D%7D%20)
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case
BaCl₂(aq) + Na₂SO₄(aq) = BaSO₄(s) + 2NaCl(aq)
Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) = BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq)
Ba²⁺(aq) + SO₄²⁻(aq)= BaSO₄(s)
Answer:
C. 500 cm' of 1.0 mol dmº magnesium sulphate solution.
Explanation:
Let us look at each of the solutions individually;
CaCl2 has three particles
K2SO4 has three particles
MgSO4 has two particles
C2H5OH has only one particle
The number of moles of moles in 250 cm of 2.0 mol dm-3 potassium chloride is 250/1000 * 2 = 0.5 moles having two particles
Also; number of moles in 500 cm' of 1.0 mol dm-3 magnesium sulphate solution= 500/1000 * 1 = 0.5 moles having two particles