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Ch.17 - Aqueous Ionic Equilibrium

Chapter 17, Problem 133

A 25.0-mL volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M hydrochloric acid for neutralization. A 10.0-mL volume of a phosphoric acid solution requires 34.9 mL of the sodium hydroxide solution for complete neutralization. Calculate the concentration of the phosphoric acid solution.

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Hi everyone here we have a question that says 17. mL of 0.212 molar hydrochloric acid was neutralized with 29.3 millimeters of a potassium hydroxide solution. The same potassium hydroxide solution was used to titrate 15. mL of an arsenic acid solution. 38.2 ml of the potassium hydroxide solution was used to neutralize the arsenic acid solution. What is the polarity of the arsenic acid solution? So we have 7.17.2 ml of hydrochloric acid times one leader Over 1000 ml Times 0.212 moles over one liter. So our middle leaders are canceling out and our leaders are canceling out. And that gives us three 646.4 Times 10 to the -3 moles of hydrochloric acid. Next we have 3. times 10 to the negative three moles of hydrochloric acid times one mole a potassium hydroxide over one mole of hydrochloric acid Times 29.3 ml over one times one liter Over 1000 ml which equals zero 0. 445 more potassium hydroxide. So now we have 0.2 male leaders potassium hydroxide times one leader Over ml Times 0.12445 moles over one liter times one mole of arsenic acid over three moles of potassium hydroxide times 15 mL, times one liter Over 1000 ml and that equals 0.106 Molar. And that is our final answer. Thank you for watching. Bye