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.
Two independent variables could change at the same time, and you would not know which variable affected the dependent variable
The correct answer to the top one is d
Answer:
it is a replacement reaction
A qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2) is given Ba²⁺(aq) + SO₄²⁻ (aq) → BaSO₄(s).
<h3>What is qualitative test?</h3>
Qualitative test measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation of the sample.
<h3>Qualitative test for sulfate in alum crystals </h3>
When an aqueous solution of a barium salt (BaCl₂) is mixed with an aqueous solution containing sulfate, a white precipitate of insoluble BaSO₄ forms according to the net ionic equation given below;
Ba²⁺(aq) + SO₄²⁻ (aq) → BaSO₄(s)
Thus, a qualitative test for sulfate in alum crystals using ionic reactions of barium chloride (BaCl2) is given Ba²⁺(aq) + SO₄²⁻ (aq) → BaSO₄(s).
Learn more about qualitative test here: brainly.com/question/2109763
#SPJ1
