A student adds 15.0 mL 2.00 M Acetic Acid and 10.0 mL 0.03 M HCl. Determine the pH of the resulting solution. Ka HC2H3O2 = 1.80E-5

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A student adds 15.0 mL 2.00 M Acetic Acid and 10.0 mL 0.03 M HCl. Determine the pH of the resulting solution. Ka HC2H3O2 = 1.80E-5

Chemistry
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the contribution of [H+] by the acetic acid alone is very considering the small dissociation of a weak acid. Anyway you can calculate the [H+] by the weak acid and calculate how many moles of H+ you have in the 15 mL (~ 0.00009 moles H+) Also you can calculate how many moles of H+ you have in the 10mL of 0.03M HCl acid. (~0.0003 mol). The amount of H+ given by the strong acid is higher than the weak acid and increasing the amount of H+ the weak acid is going to dissociate less than being alone and the amount of H+ given to the mix by the weak acid is going to be even less than 0.00009 moles. Besides the volume is change and the concentration of the weak acid will decrease You can calculate the pH with the value of the concentration of the strong acid considering the dilution in the change of volume. Calculate the new concentration of the 10 mL 0.03M HCl in 25mL final volume, and calculate the pH with that concentration. I got pH 1.92 https://www.youtube.com/watch?v=UGs0hrIkjss
How can you use hasselbalchs equation when there's 2 acids @Cuanchi

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you can not use hasselbach is for buffers when you have a weak (acid or base) and the salt of the weak. Can be used also in titrations of a weak with an strong before the equivalence point

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