When 14.7 mL of aqueous HBr (a strong acid) was added to water, 0.482 L of a sol

Question

When 14.7 mL of aqueous HBr (a strong acid) was added to water, 0.482 L of a solution with a pH of 4.23 was produced. What was the molarity of the original HBr solution?

- Do I find molarity from pH? then use the M1V1=M2V2? If so, could you go over how to get molarity and if not, please explain steps clearly.

Thank you!
When 14.7 mL of aqueous HBr (a strong acid) was added to water, 0.482 L of a solution with a pH of 4.23 was produced. What was the molarity of the original HBr solution?

- Do I find molarity from pH? then use the M1V1=M2V2? If so, could you go over how to get molarity and if not, please explain steps clearly.

Thank you!

Explanation / Answer

HBr => H+ + Br-

[H+] = 10^(-pH) = 10^(-4.23) = 5.8884 x 10^(-5) M


Moles of HBr = moles of H+ = volume of diluted solution x [H+]

= 0.482 x 5.8884 x 10^(-5) = 2.838 x 10^(-5) mol


Original volume of HBr = 14.7 mL = 0.0147 L


Molarity = moles/original volume of HBr

= 2.838 x 10^(-5)/0.0147

= 0.00193 M = 1.93 x 10^(-3) M

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