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.
Sunlight, soil, water, leaves,
Endothermic reactions are reactions that require heat in the course of the process. The heat of reaction in this case is positive which means the energy of the products is greater than the energy of the reactants. In this case, the answer to this problem would have to be <span>1. must be greater than the required ΔH </span>
85-12 = 73 degrees needed
4.186 J/degree Celsius, so
73 degrees * 4.186 J/degree = 305.578 J to raise 1 gram 73 degrees
there are 675 grams, so 305.578 * 675 = 206265.15 J
2.06 x 10^5 J are needed
Answer:
true
Explanation:
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