Calculate the pH of each of the following strong acid solutions.
0.697 g of HClO3 in 2.97 L of solution. so i convenrted grams to moles then moles/L to get M of HCl03 then i did -log of that cocentration and put htat as the ph. did i do something wrong? i got 2.415 but it was wrong please help!
a similar problem i am having troube with is 15.10 mL of 2.00 M HCl diluted to 0.530 L
When you don't show your work we can't know what you did wrong. Next time show your work and we can spot the problem almost immediately.
You must be pushing the wrong keys.
0.697/84.459 = ?
?/2.97 = 0.002779
and -log = 2.556
To calculate the pH of a strong acid solution, you are on the right track. However, let's go through the steps in more detail to identify any mistakes.
Step 1: Convert grams to moles.
Given that you have 0.697 g of HClO3, you need to determine the number of moles. To do this, divide the mass by the molar mass of HClO3. The molar mass of HClO3 is calculated as follows:
H (1.01 g/mol) + Cl (35.45 g/mol) + 3O (16.00 g/mol) = 82.46 g/mol
Moles of HClO3 = 0.697 g / 82.46 g/mol = 0.00845 mol (rounded to five decimal places)
Step 2: Calculate molarity (M) of the solution.
Molarity is defined as moles of solute per liter of solution (mol/L). In this case, we have 0.00845 mol of HClO3 in 2.97 L of solution, so:
Molarity (M) = 0.00845 mol / 2.97 L = 0.00284 M (rounded to five decimal places)
Step 3: Calculate the pH of the solution.
The pH of a strong acid is directly related to the concentration (molarity) of the acid, according to the equation:
pH = -log[H+]
Since HClO3 is a strong acid, it completely dissociates in water, so the concentration of H+ ions is equal to the concentration of HClO3.
pH = -log(0.00284) = 2.5453 (rounded to four decimal places)
Therefore, the pH of the strong acid solution with 0.697 g of HClO3 in 2.97 L of solution is approximately 2.5453.
Please recheck your calculations to see if you made any errors along the way.