<span>NaCl (halite) or KCl (sylvite) </span>
Answer:
The basicity of HCOOH (otherwise known as formic acid) is 1
Explanation:
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
Answer:
H₃PO₄ → 3H⁺ + PO₄³⁻
CaSO₄ → Ca²⁺ + SO₄²⁻
b. CaCl₂
Explanation:
When H₃PO₄ is dissolved in water, there are produced the H⁺ and PO₄³⁻ ions. The equation is:
H₃PO₄ → 3H⁺ + PO₄³⁻
In the same way, CaSO₄ is dissolved in:
CaSO₄ → Ca²⁺ + SO₄²⁻
b. Now, in a reaction of an acid (HCl) and a base (Ca(OH)₂), water, H₂O and a salt are produced:
2 HCl + Ca(OH)₂ → 2H₂O + Salt
The ions that are not present in the reaction are Cl⁻ and Ca²⁺, the salt is CaCl₂ and the balanced reaction is:
2 HCl + Ca(OH)₂ → 2H₂O + CaCl₂
