A buffer was made by mixing 20cm3 of 0.05M NaOH and 20cm3 of 0.25M propanoic acid. Calculate the pH. Ka=1.34x10^-5

pH is calculated using the equation pH=-log[H+]. The concentration of H+ ions, therefore, needs to be calculated. The amount of each substance can be calculated by multiplying the concentration by the volume of solution; Mol NaOH = 0.05x20/1000 = 0.001 Mol Propanoic Acid = 0.25x20/1000 = 0.005 When the two solutions are mixed, the NaOH reacts with the propanoic acid to form a propanoate ion and water as follows; NaOH + CH3CH2COOH => CH3CH2OO-Na+ + H2O The concentration of propanoate ions and propanoic acid in the mixture can therefore be calculated; [Propanoate ion] = [NaOH] = 0.001x1000/40 = 0.025M [Propanoic acid] = (0.005-0.001)x1000/40 = 0.1M The acid dissociation constant can now be used to find the concentration of H+ ions Ka = [H+][A-]/[HA]
where HA is the acid and A- is the propanoate ion. Rearranging gives: [H+] = Ka[HA]/[A-] Inserting values into the equation gives [H+] = 5.36x10^-5 Using the equation for pH gives pH = 4.27

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Answered by Michael C. Chemistry tutor

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