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maks197457 [2]
3 years ago
6

What is the kw of pure water at 50.0°c, if the ph is 6.630? 5.50 × 10-14 2.13 × 10-14 1.00 × 10-14 2.34 × 10-7 there is not enou

gh information to calculate the kw?
Chemistry
2 answers:
dimaraw [331]3 years ago
8 0

Answer:

The K_w value of water at 50.0° C is 5.50\times 10^{-14}.

Explanation:

The pH of water at 50.0° C = 6.630

pH=-\log[H^+]

6.630=-\log[H^+]

[H^+]=2.344\times 10^{-7}M

Since, water is a neutral compound wuith equal number of hydrogen ions and hydroxide ions.

[OH^-]=[H^+]=2.344\times 10^{-7}M

H_2O\rightleftharpoons H^++OH^-

The expression for an ionic product of water is given as:

K_w=[H^+][OH^-]

Substituting the values:

K_w=[2.344\times 10^{-7}M][2.344\times 10^{-7}M]=5.49\times 10^{-14}\approx 5.50\times 10^{-14}

The K_w value of water at 50.0° C is 5.50\times 10^{-14}.

Andreyy893 years ago
4 0
Answer is: Kw of pure water at 50.0°C is <span>5.50 × 10-14</span><span>.
pH = 6.630.
pH = -log</span>[H⁺].
[H⁺] = 10∧(-pH).
[H⁺] = 10∧(-6.63) = 2.34·10⁻⁷ M.
[H⁺] · [OH⁻] = x.
Kw = ?.<span>
Kw = [H</span>⁺] · [OH⁻]. <span>
Kw = x².
Kw = (</span>2.34·10⁻⁷ M)².
Kw = 5.50·10⁻¹⁴ M².
Kw is ionic product of water.

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