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:
Ok so the answer for 9 is
x/6=4
x=24
Explanation:
Solve for x by simplifying both sides of the equation, then isolating the variable.
Answer:
compounds are elements include an element and a compound
Explanation:
elements in the decomposition reaction is the substance that cannot be separated into simpler substances. Compounds, technically act as a reactant in the decomposition reaction, but since the reaction breakdown one substance into two or more, sometimes it exists in the product
Mass box C is 10+5. (So C is 15)
But if C was 30, how many times could you put B (5) into it?
30/5 = 6
You would need 6 boxes of B to make 30 grams of C.
