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mezya [45]
3 years ago
8

A 1.0L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. Calculate the p

H of the solution upon the addition of 10.0 mL of 1.00 MHCl to the original buffer.
Express the your answer to two decimal places.
***will mark Brainliest for right answer ***
Chemistry
1 answer:
Crazy boy [7]3 years ago
7 0
Starting pH = pKa because the acid and conjugate base have equal concentrations.
pH = pKa = -log(1.8*10–5) = 4.74

0.0100 L • 1.00 M = 0.0100 mol HCl, which reacts with the conjugate base, producing 0.0100 mol of weak acid.

You now have 0.110 mol weak acid and 0.090 mol conjugate base. Use the Henderson-Hasselbach equation to solve for pH now.

pH = 4.74 + log(0.090/0.110) = 4.65

The equation technically says to use concentrations of the acid and base, but that doesn’t actually matter, because it would give the same answer.
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N204(0) + 2NO2(g)
user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

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\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}

Comparing Q with K( the equilibrium constant) :

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Q is the product of the ion ions from the reacting substance  

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

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Setup 1 :

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\tt Q=\dfrac{0.230^2}{0.420}=0.126

Q>K⇒The reaction moved to the left (reactants)

8 0
2 years ago
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A is correct <em>answer.I </em><em>hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>you</em>

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