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:
0.33 cal⋅g-1°C-1
Explanation:
The amount of heat required is determined from the formula:
q= mcΔT
To see more:
https://api-project-1022638073839.appspot.com/questions/what-is-the-specific-heat-of-a-substance-if-1560-cal-are-required-to-raise-the-t#235434
CH3CH2OH = 46.068 (molar mass)
1.50g(1 mole / 46.0680) =
0.03256 moles
In a 0.01 M solution of HCl, Litmus will be red. Litmus paper will turn into red in acidic conditions. Hydrochloric acid is an acid. Litmus is an indicator for acidity and alkalinity made from inchens.
I think is 1 and a half km
