How to calculate pH of HCl in water?

Calculate the pH of 0.1825 g of HCl dissolved in 200 mL of distilled water.?

  • Answer:

    First you must convert this HCl value into mols, then into mol/L. HCl molar mass: 36.46 g/mol. 0.1825 g HCl x 1 mol HCl/ 36.46 g HCl = 5.005*10^-3 mol HCl Convert mL into L = 200 mL x 1 L/1000 mL = 0.200 L 5.055*10^-3 mol HCl/ 0.200 L = 2.50*10^-2 mol/L HCl Now you have to put it into an equation. pH is equal to -log of the concentration of hydronium (H30+) in a mixture. But in this case, we have HCl; HCl is a strong acid, and completely dissociates into its constituent elements, and doesn't form a equilibrium like a weak acid would. HCl (aq) --> H+ + Cl- Therefore, the concentration of HCl = concentration of Hydronium. pH = -log[H30] pH = -log(2.50*10^-2) pH = 1.60 The pH is 1.60 which means it's acidic, obviously since HCl is an acid.

Super at Yahoo! Answers Visit the source

Was this solution helpful to you?

Other answers

First you must convert this HCl value into mols, then into mol/L. HCl molar mass: 36.46 g/mol. 0.1825 g HCl x 1 mol HCl/ 36.46 g HCl = 5.005*10^-3 mol HCl Convert mL into L = 200 mL x 1 L/1000 mL = 0.200 L 5.055*10^-3 mol HCl/ 0.200 L = 2.50*10^-2 mol/L HCl Now you have to put it into an equation. pH is equal to -log of the concentration of hydronium (H30+) in a mixture. But in this case, we have HCl; HCl is a strong acid, and completely dissociates into its constituent elements, and doesn't form a equilibrium like a weak acid would. HCl (aq) --> H+ + Cl- Therefore, the concentration of HCl = concentration of Hydronium. pH = -log[H30] pH = -log(2.50*10^-2) pH = 1.60 The pH is 1.60 which means it's acidic, obviously since HCl is an acid.

Tulpe

Related Q & A:

Just Added Q & A:

Find solution

For every problem there is a solution! Proved by Solucija.

  • Got an issue and looking for advice?

  • Ask Solucija to search every corner of the Web for help.

  • Get workable solutions and helpful tips in a moment.

Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.