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belka [17]
3 years ago
8

What is the pH of a solution after the addition of 30.0 mL of 0.100 M NaOH to 50.0 mL of 0.10 M HBr?

Chemistry
1 answer:
Pani-rosa [81]3 years ago
7 0

Answer:

pH = 1.6

Explanation:

  • HBr + NaOH ⇒ NaBr + H2O

0 mL NaOH:

  • HBr ↔ H3O+  +  Br-

⇒ [ H3O+] = M HBr = 0.1 M

⇒pH = -log [H3O+] = 1

30 mL NaOH:

⇒ mol NaOH = 0.1 mol / L * 0.03 L =  3 E-3 mol

⇒ mol HBr = 0.05 L * 0.1 mol/L = 5 E-3 mol

⇒ M HBr = ( 5 E-3 mol - 3 E-3 mol) / 0.08 L = 0.025 M

⇒ pH = - log (0.025) = 1.6

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A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at
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Explanation:

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Now, volume of HNO_{3} present will be calculated as follows.

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Therefore, the total volume will be the sum of the given volumes as follows.

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or,               = 0.2991 L

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     CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}

As,      k_{a} = \frac{k_{w}}{k_{b}}        

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Now,   [HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}

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                      = 2.546 \times 10^{-6}

Now, pH will be calculated as follows.

              pH = -log [H_{3}O^{+}]

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