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 ![\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
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
Answer:
Solutions in water are especially common, and are called aqueous solutions. Non-aqueous solutions are when the liquid solvent involved is not water.
Answer would be A)the atom
Answer:
<h3>The answer is 1.30288 × 10³ g</h3>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume = 95.8mL
density = 13.6g/mL
We have
mass = 13.6 × 95.8 = 1302.88
We have the final answer as
<h3>1.30288 × 10³ g</h3>
Hope this helps you
Answer:
B. Equal amounts of all gases have the same volume at the same
conditions
Explanation:
Amedo Avogadro found the relationship between volume of a gas and the number of molecules contained in the volume.
The law states that "equal volumes of all gases at the same temperature and pressure contains equal number of molecules or moles".
The law describes the behavior of gases when involve in chemical reactions. It enables one to change over at will in any statement about gases from volumes to molecules and vice versa.
So, the right option is B which implies that equal amounts of all gases have the same volume at the same conditions.