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fiasKO [112]
3 years ago
9

Consider the reaction: 2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(l); H = –118 kJ. Calculate the change in temperature when 50.

0 mL of 0.20 M HCl(aq) and 50.0 mL of 0.10 M Ba(OH)2(aq) are mixed together. The density ofthe solution is 1.00 g/mL and Solution= 4.184 J/g-°C
Chemistry
1 answer:
Alborosie3 years ago
6 0

Answer:

ΔT = 1,41°C

Explanation:

For the reaction:

2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2 H₂O(l); ΔH = –118 kJ.

If moles of reaction are:

HCl:

0,0500L×0,20M = 0,01moles

Ba(OH)₂:

0,0500L×0,10M = 0,005moles

The heat produced is:

0,005mol×-118 kJ/mol = -0,59 kJ = -590 J

The temperature increasing is:

Q = -C×m×ΔT

Where Q is heat (-590 J), m is mass of solution (100mL = 100g -density of 1,00g/mL) and ΔT is change in temperature.

Replacing:

-590J = -4,184J/g°C×100g×ΔT

<em>ΔT = 1,41°C</em>

<em></em>

I hope it helps!

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<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

11=-\log [OH^-]

[OH^-]=1.0\times 10^{-11}M

Thus, the OH^- concentration is, 1.0\times 10^{-11}M

<u>(b) a solution of 0.10 M NH_3</u>

As we know that 1 mole of NH_3 is a weak base. So, in a solution it will not dissociates completely.

So, the OH^- concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12=-\log [OH^-]

[OH^-]=1.0\times 10^{-12}M

Thus, the OH^- concentration is, 1.0\times 10^{-12}M

<u>(d) a solution of 0.10 M NaOH</u>

As we know that NaOH is a strong base. So, it dissociates to give Na^+ ion and OH^- ion.

So, 0.10 M of NaOH in a solution dissociates to give 0.10 M of Na^+ ion and 0.10 M of OH^- ion.

Thus, the OH^- concentration is, 0.10 M

<u>(e) a 1\times 10^{-4}M solution of HNO_2</u>

As we know that 1 mole of HNO_2 in a solution dissociates to give 1 mole of H^+ ion and 1 mole of NO_2^- ion.

So, 1\times 10^{-4}M of HNO_2 in a solution dissociates to give 1\times 10^{-4}M of H^+ ion and 1\times 10^{-4}M of NO_2^- ion.

The concentration of H^+ ion is 1\times 10^{-4}M

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (1.0\times 10^{-4})

pH=4

Now we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

10=-\log [OH^-]

[OH^-]=1.0\times 10^{-10}M

Thus, the OH^- concentration is, 1.0\times 10^{-10}M

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

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