What concentrations of CH3COOH and CH3COONa are needed to prepare a 0.10M buffer at pH 5.0?
From Henderson's equation we know for an acid bufferpH= pKa+ log (salt/acid)We know, Ka of CH3-COOH= 1.8 x 10^ -5.So pKa= 4.7447;If pH of the buffer should be 5.00 thenlog(CH3-COONa/CH3-COOH)= 5.0- 4.7447 = 0.2553.So the ratio of the concentration of Sodium acetate to Acetic acid should be= 10^ 0.2553 =1.8001~ 1.8.So if the concentration of Sodium acetate is 1.8 times more Concentrated than acetic acid, then the pH of the solution will be 5.0 !!!For example if the concentration of CH3COOH is 1.0 M, then if the concentration of CH3-COONa is 1.8 M the pH of the solution will be 5.00 !!!
What's the pH of a solution containing 0.1 moles of H₂SO₄ and 0.01-mole KOH in 2l of water?
The reaction between KOH and H2SO4 is : 2 KOH + H2SO4 = K2SO4 + H2O2 moles of KOH would neutralise 1 mole of the acid.Hence, 0.01 mole of KOH will react with .005 mole of H2SO4.No. of moles of H2SO4 remaining after neutralisation of KOH = 0.1-0.005 = 0.095.K2SO4 is a salt of a strong acid and strong base, and hence it will not affect the pH of the mixture.The pH of the solution will depend on the concentration of the acid only.No. of moles the acid in 2 litre of the solution = 0.095 Concentration of the acid = 0.095/2= 0.0475 M pH = -log[H+] = -log[4.75 x 10^-2] = +2 x log10 - log[4.75] = +2 - 0.6766 = 1.32 (answer).
A buffer solution is prepared by adding 20 ml of 0.1M NaOH solution to 40ml of 0.2M CH3COOH solution. What is the pH of the buffer?
There are a couple of pieces to this puzzle. Hopefully, they are arranged below such that they make sense.First, what reaction is going to be driving the pH? There is some acetic acid (AcOH) and some sodium hydroxide (NaOH) which will react with the AcOH and convert some of it to sodium acetate (AcONa). So what will be present in the buffer is AcOH and AcONa, assuming that the NaOH is not in excess of the AcOH (more on this in a bit). The reaction of interest is:AcOH = AcONa + H+This is an equilibrium defined by Ka (1.74*10^-5) and:Ka = [AcONa][H+]/[AcOH]pH is defined bypH = -log[H+]So, rearranging Ka so that [H+] is solved for gives[H+] = Ka[AcOH]/[AcONa]and thereforepH = -log(Ka[AcOH]/[AcONa])Next, what are the concentrations of AcOH and AcONa? The 20 mL of 0.1 M NaOH translates to 0.02 L * 0.1 mol/L = 0.002 mol NaOH. The 40 mL of 0.2 M AcOH translates to 0.04 L * 0.2 mol/L = 0.008 mol AcOH. There is more AcOH than NaOH so all of the NaOH will be consumed by the AcOH and AcONa (0.002 mol) will be formed. In this process, the amount of AcOH will be decreased by the amount of AcONa formed - AcOH = 0.008–0.002 = 0.006 mol.Next, what are the concentrations of AcOH and AcONa? First, note that the total volume is now 60 mL or 0.06 L. Using this volume the concentrations can be calculated:[AcOH] = 0.006 mol/0.06 L = 0.1 M[AcONa] = 0.002 mol/0.06 L = 0.033 MNow, put these numbers in for the equation defining pH and solve:pH = -log(1.74*10^-5)(0.1)/(0.033) = -log(5.21*10^-5) = 4.28So, long story short, the pH will be about 4.28.