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:
P₂ = 140 KPa
Explanation:
Given data:
Initial volume = 8.0 L
Final volume = 4.0 L
Initial pressure = 70 KPa
Final pressure = ?
Solution:
According to Boyle's law
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 70 KPa ×8.0 L/4.0 L
P₂ = 560 KPa .L / 4.0 L
P₂ = 140 KPa
Answer:
Yes, because the light was the manipulated variable
Explanation:
A saturated is one in which the atoms are linked by single bonds. :)