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How can 80mL of HCl solutions of pH 2 and 3 be obtained from a solution of 0.2M HCl? (Hint: make 100mL solutions using 100mL volumetric flask)

PLS chemistry pros answer this, I need the answer of this in less than 6 hours :( I got an exam and me and my pals are having hard time with this question

Respuesta :

Edufirst
Case 1: pH = 2.

1) find the molar concentration of H3O + or H+

pH = - log [H +] = 2 => [H+] = 10 ^-2 = 0.01 M

2) find the number of moles that will have 80 ml of a 0.01 M solution:

M = n /V => n = M * V = 0.01 M * 0.080 liter = 0.00080 moles

3) Find the volume of the 0.2 M solution that has 0.00080 moles of H+

 M = n / V => V = n / M = 0.00080 mol / 0.2 M = 0.004 liter = 4 ml.

That means that you can take 4 ml of the solution 0.2M HCl (which will contain 0.00080 moles of H+) and add water until complete 80 ml.  

You can verify: 0.00080 mol / 0.080 liter = 0.01 M

pH = log (1 / 0.01) = 2 .

Case 2: pH = 3.

1) concentration of H+

pH = - log [H+] = 3 => [H+] = 10^ -3 = 0.001

2) number of moles that will have 80 ml 0.001 M solution

M = n / V => n = M * V = 0.001M * 0.080 liter = 0.000080  moles

3) Volume of the 0.2 M solution that will contain 0.000080 moles

M = n / V => V = n / M = 0.000080 mol / 0.2 M = 0.0004 liter = 0.40 ml

Then, you can take 0.40 ml of 0.2 M HCl solution and add water until complete the 80 ml.

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