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Vanyuwa [196]
2 years ago
11

What volume of 3.00 M HCl in liters is needed to react completely (with nothing left over) with 0.750 L of 0.100 M Na2CO3?

Chemistry
1 answer:
Alla [95]2 years ago
5 0

Answer:

0.0500L of 3.00M HCl are needed to react completely

Explanation:

Hydrochloric acid HCl reacts with sodium carbonate, Na2CO3, as follows:

2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂

<em>Where 2 moles of HCl reacts per mole of Na₂CO₃</em>

<em />

To find the volume of 3.00M HCl needed to react completely we need first to calculate the moles of Na₂CO₃ added and, with this chemical equation we can solve for moles of HCl as follows:

<em>Moles Na₂CO₃:</em>

0.750L * (0.100moles Na₂CO₃ / L) = 0.0750moles

<em>Moles HCl:</em>

0.0750moles Na₂CO₃ * (2 moles HCl / 1 mole of Na₂CO₃) = 0.150 moles of HCl

<em>Volume of 3.00M HCl:</em>

0.150 moles of HCl * (1L / 3.00moles) =

<h3>0.0500L of 3.00M HCl are needed to react completely</h3>
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Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found
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Answer:

The concentrations are :

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[Asc^{2-}]=5.92\times 10^{-8} M

The pH of the solution is 3.15.

Explanation:

H_2Asc\rightleftharpoons HAs^-+H^+         K_{a1}=1.0\times 10^{-5}

Initial

c                0              0

Equilibrium

c-x                x          x

K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}

1.0\times 10^{-5}=\frac{x\times x}{(c-x)}

1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}

Solving for x:

x = 0.000702 M

[HAsc^-]=0.000702 M

HAsc^-\rightleftharpoons As^{2-}+H^+        K_{a2}=5\times 10^{-12}

Initially

x                0          0

At equilibrium ;

(x - y)            y         y

K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}

5\times 10^{-12}=\frac{y\times y}{(x-y)}

5\times 10^{-12}=\frac{y^2}{(x-y)}

Putting value of x = 0.000702 M

5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}

y=5.92\times 10^{-8} M

[Asc^{2-}]=5.92\times 10^{-8} M

Total concentration of [H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M

The pH of the solution :

pH=-\log[H^+]

pH=-\log[7.0206\times 10^{-4} M}=3.15

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