Question

Please answer part 1 and 2

Part 1:

A chemist dissolves 709. mg of pure barium hydroxide in enough water to make up 90. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits. x 6 ?

Part 2:

Answer 1

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Part 1 mass of Bacot), = 709 mg 0.7099 (1000 mg - 19) molar mals of Balot/2 = 171.34 g/mol mules of Balot = mass - 0.7092 molar man 171.34glmol volume of solution - 90 mL = 0.092 = = 4.14 X 10-3 mol (1000 m L=12) concentration of Balot = mole) volume = 4.14 8 10-3 mol - 0.046 M 0.092 , so, It dissociate completely into its lons in Baloth, strong base aqueous soleation. 0.046 M Becom), (49) Ba²+ (aq) & 2014 lagl After dwsociation 2x0.046M = 0.092 M [OH-] = 0.092M Þohe - log [OH-] = -log 10.092) = 1.04 At 25 C pH = 14- pot = 14-1.04 = 12.96 I pH= 12.96 Part 2. pka -log ka the value of ka. oo more is Higher pkg ka to bka the acidity of an acid Relative spength Les acidic Least andre ko Acid 10.3 4.80 X10-11 7.94 x 10-12 4.7X10-9 8.3 most acidic Relatue shepth of werle acid DAYB