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Brrunno [24]
4 years ago
15

Write equations for the reaction of 1 mol of Barium Hydroxide with 2 Moles of hydrochloric acid

Chemistry
1 answer:
gtnhenbr [62]4 years ago
5 0

Answer:

Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l).

Explanation:

1 mol of Barium Hydroxide with 2 moles of hydrochloric acid  to produce 1 mol of barium chloride and 2 moles of water according to the equation:

<em>Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l). </em>

<em></em>

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
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<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

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