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3 years ago

Pure water has a pH of 7. Pure water _______.

Chemistry

1 answer:

raketka [301]3 years ago

8 0

Answer:

The concentration of Hydrogen Cation is:

Pure water is NEUTRAL

Explanation:

pH = It is negative logarithm of activity of Hydrogen ions.

$pH = - log(a_{[H^{+}]})$

Activity = concentration for dilute solutions

There are three types of solution based on pH:

1. Acidic solution : Those solution which have pH between 0 to 7 or less than 7 are acidic in nature . They turns blue litmus red.They contain more H+ ions than OH- ions

2. Basic : Those solution which have pH value greater that 7 of between 7 to 14 are basic in nature . They turn red litmus blue.They contain more OH- ions then H+ ions

3 . Neutral : Those solution which have pH = 7 are neutral . They are neither acidic nor basic in nature .They have equa;l concentration of H+ and OH- ions in the solution.

The given water has pH = 7 .This means it is neutral and has low H+ ion concentration in solution. The H+ ion are :

$10^{-7}$ M in water.

OH- =

$10^{-7}$ M in water

So overall neutral

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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Answer:

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