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.
The correct answer is C. because it could lead to an increase in ocean levels
Diffusion, facilitated diffusion, filtration, and osmosis
Answer:
-The stronger electrostatic forces of attraction between the oppositely charged ions causes the Sodium chloride to break apart until it completely dissolves in the water.
Explanation:
-Sodium Chloride has positively charged sodium ions, 
and negatively charged chloride ions, 
.
-Water on the other hand has positively charged Hydrogen ions,
and negatively charged Oxygen ions, 
due to the difference in electroneganivity.
-When dissolved in water, the positively charged sodium ions will attract the partially negatively charged oxygen ions. The negatively charged chloride ions will be attracted to the positively charged hydrogen ions in the reaction as below:
