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boyakko [2]
3 years ago
6

. A 500.0 mL buffer solution contains 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of thi

s solution be after the addition of 20.0 mL of 1.00 M HCl solution
Chemistry
1 answer:
julia-pushkina [17]3 years ago
7 0

Answer:

pH = 4.57

Explanation:

pH = pKa + log ([OAc⁺]/[HOAc])

Ka(HOAc) 1.8 x 10⁻⁵ => pKa = -log(1.8 x 10⁻⁵) = 4.74

[OAc⁻] = 0.20M

[HOAc] = 0.30M

pH = 4.74 + log([0.20]/[0.30]) = 4.47 + (-0.17) = 4.57

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The sugar would dissolve better under warmer temperatures because the hot particles interfere with the sugars particles.
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Ammonia is a weak base because it produces ___ ions in solution
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Only a few ions (in comparison to others that totally break apart)
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Can someone name an element that doesn’t match its abbreviation?
sp2606 [1]

Answer:

Gold, silver and sodium.

Explanation:

Gold, silver and sodium are the name of the elements which does not match to their abbreviation. It is because those names are comes from other language. For example, the symbol of sodium is Na which comes from Natrium that is a Latin word. The symbol of gold is Au which comes from Aurum that is also a Latin word. The symbol of silver is Ag which comes from a Latin word i. e. Argentum.

5 0
4 years ago
Citric acid, also known as vitamin C, is a triprotic acid with the formula H3C6H5O7. The pKa values are 3.13, 4.76, and 6.40. Wh
Ronch [10]

Answer:

pH = 5.58

Explanation:

Given that :

Citric acid  is a triprotic acid with the formula H_3C_6H_5O_7 . Tripotic acids are acids that have the tendency to yield three protons per molecule during the phases of dissociation, therefore they have three pka.

For Citric acid; the pka values are:

pKa _1 = 3.13\\pKa_2 = 4.76\\pKa_3 = 6.40

So, the pH of sodiummonohydrogencitrate with the chemical formula Na_2HC_6H_5O_7 which can act as an acid and a base (i.e it is said to be amphoteric) can be calculated as :

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Therefore, the approximate pH of a solution of sodium monohydrogencitrate = 5.58

5 0
3 years ago
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Valentin [98]
According to Raoult's low:
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∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr 
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808 
when we assume the moles of solute NaCl = X 
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
 So X (the no.of moles of NaCl) = 0.027 m
8 0
3 years ago
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