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1 year ago

Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH:

Chemistry

1 answer:

ale4655 [162]1 year ago

3 0

Answer:

(a) 7.0 (b) 7.30 (c) 8.42

Explanation:

(a) HBr is a strong acid and NaOH is also a very strong base, hence, the solution will be neutral give a pH of 7.0

(b) HClO3

ClO2- + H20 <=> HClO2- + OH-

ClO2- = 0.1*0.08/0.18 = 0.044M

Kb=Kw/Ka=1.0 x 10^-14/1.1 x 10^-2=9.09x10^-13

[OH-] = 2.0 x 10^-7M

p[OH] = 6.7

pH = 7.3

C6H5COO- + H20 <=> C6H5COOH + OH-

[C6H5COO-] = 0.1 * 0.80/0.180 = 0.44

Kb=Kw/Ka=1.0 x 10^-14/6.3 x 10-²=1.59x10^-10

[OH] = 1.59x10-10/0.44 = 2.69x10-6

pOH = 5.58

pH = 8.42

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1 year ago

Determine the molar solubility of CuCl in a solution containing 0.050 KCl. Ksp of CuCl is 1.0 x 10-6

Volgvan

Answer:

$2.0x10^{-5}\frac{mol}{L}$

Explanation:

Hello!

In this case, since the dissolution of copper (I) chloride is:

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We can represent the molar solubility via the reaction extent as $x$ , however, since there is 0.050 M KCl we immediately add such amount to the chloride ion concentration since KCl is readily ionized; therefore we write: