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kakasveta [241]
3 years ago
12

Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.

Chemistry
1 answer:
velikii [3]3 years ago
6 0
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
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Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.
baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that M_{1}V_{1} = M_{2}V_{2}

Where,    V_{1} = total volume

               V_{2} = initial volume

Therefore, new concentration of CH_{3}COOH = \frac{M_{2}V_{2}}{V_{1}}

                                        = \frac{60 \times 0.5}{70}

                                        = 0.43 M

New concentration of NaOH = \frac{M_{2}V_{2}}{V_{1}}

                                               = \frac{10 \times 1.0}{70}

                                               = 0.14 M

So, the given reaction will be as follows.

              CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O

Initial:             0.43          0.14                     0

Change:          -0.14        -0.14                    0.14

Equilibrium:    0.29          0                       0.14

As it is known that value of pK_{a} = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

           pH = pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

                 = 4.74 + log \frac{0.14}{0.29}

                 = 4.74 + (-0.316)

                 = 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

6 0
3 years ago
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