The change in pH of a 1.00 L of a buffered solution preparing by mixing 0.50 M acetic acid (Ka = 1.8 x 10^-5) and 0.50 M sodium acetate when 0.010 mole of NaOH is added is 4.75
when the same amount 0.010 mole of NaOH was added to 1.00 L of water the pH = 12
Explanation:
given that:
concentration of acetic acid = 0.50 M
Concentration of base sodium acetate = 0.50 M
ka = 1.8 x 10^-5)
pka = -log [ka]
pka = 4.74
From Henderson-Hasselbalch Equation:
pH = pKa + log ![\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
pH = 4.74 + Log ![\frac{[0.5]}{[0.5]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0.5%5D%7D%7B%5B0.5%5D%7D)
pH = 4.74 + 0
pH = 4.74
Number of moles of NaOH = 0.010 moles
volume 1 litre
molarity = 0.010 M
Moles of acetic acid and sodium acetate before addition of NaOH
FORMULA USED:
molarity = 
acetic acid,
0.5 = number of moles
0.5 is the number of moles of sodium acetate.
number of moles of NaOH 0.010 moles
NaOH reacts in 1:1 molar ratio with acetic acid so
number of moles in acetic acid = 0.5 - 0.010 = 0.49
number of moles in sodium acetate = 0.5 +0.010 = 0.51
new pH
pH = pKa + log
pH= 4.74 + log[0.51] - log[0.49]
pH= 4.75
PH of NaOH of 0.01 M (BASE)
pOH = -Log[0.01]
pOH = 2
pH can be calculated as
14= pH +pOH
pH= 14-2
pH = 12
"High temperatures make the gas molecules move more quickly" is the one sentence among all the choices given in the question that most likely explains why this reaction is carried out at high temperature. The correct option among all the options that are given in the question is the third option or option "C".
Explanation:
Molar mass
The mass present in one mole of a specific species .
The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .
(a) S₈
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
Molar mass of S₈ = 8 * 32 g/mol. = 256 g/mol.
(b) C₂H₁₂
Molar mass of of the atoms are -
Hydrogen , H = 1 g/mol
Carbon , C = 12 g/mol
Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol
(c) Sc₂(SO₄)₃
Molar mass of of the atoms are -
sulfur, S = 32 g/mol.
oxygen , O = 16 g/mol.
scandium , Sc = 45 g/mol.
Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol
(d) CH₃COCH₃ (acetone)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol
(e) C₆H₁₂O₆ (glucose)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.
