What is the pH of 1M HCl + (0.1M CH3COOH and 0.1M CH3COONa-)?
H^+ + A^- => HA I: 0.001 mol H^+ 0.01 mol HA 0.01 mol A^- === C: 0.011 mol HA 0.009 mol A^- where V(tot) = 101 mL then buffer solution: pH = pKa -lg(HA/A^-) = 4.74 - lg(0.011/0.009) = 4.65
PH of CH3COOH + HCl Solution?
My solution contains 5ml of 0.1M CH3COOH and 5ml of 0.1M HCl, and I need to calculate the pH. I understand that it will be more acidic with the addition of the HCl, and that the majority of the H+ concentration comes from the strong acid now instead of the weak CH3COOH acid. Here's my work so far: .......[CH3COOH] <-> [H+] + [CH3COO-] ....I.......0.05..................0...... Added.......................0.05...\ ...C......+0.05-x..........-.05+x........ ...E......0.1+-x.............x........... Ka=1.8e-5=x^2/0.1 x=0.00134 pH=-log[0.00134]=2.87 I expected the value to be much lower than that. When I calculated the pH of 5ml H20 and 5ml .1M CH3COOH, I got 3. (I set up an ice table, 1.8e-5=x^2/0.05, x=0.00095, pH=-log[0.00095]=3). I think I am setting up the table wrong. This is from a lab, which is ahead of our lecture and the manual uses a different book than our lecture. I couldn't find any examples to verify my work. I am not really sure how the common-ion effect changes the ice table. Should I be using the Ka of HCl?
What is the pH of solution when 0.1 M CH3COOH and 0.1 M NaOH are mixed?
Here we need volume of acid CH3COOH and base NaOH. If volumes are not known we can't find pH of the resultant solutions.If we take both acids and bases are of equal volumes of V each. So the solution act as the salt solution for finding pH we need to apply salt hydrolysis formula for finding pH.Now the mixture will acts as a salt CH3COO-Na+ of week acid CH3COOH and strong base NaOH.Then pH is 7 + 0.5*(pKa+ logX) here X is concentration of salt.We known that pKa of CH3COOH is 4.75Given concentration of acid is 0.1M and base is 0.1 MNow both are equal volumes and concentrations.Let's assume concentration of salt is XNow 2VX = V(0.1+0.1)X = 0.2/2X= 0.1Now pH = 7 + 0.5*(pKa+ logX)pH = 7 + 0.5*(4.75 + log 0.1)= 7 + 0.5*(4.75–1)= 7 + 0.5*(3.75)= 7 + 1.875= 8.875.Therefore pH of the resultant solution is 8.875
Calculate the pH of a solution that is 0.15 M CH3COOH and 0.75 M CH3COONa?
O.K . First, establish the equilibrium CH3COOH <---> CH3COO- + H+ (This is a weak acid, so there will be very little dissociation: the Ka value attests to that). Also, note that CH3COONa will dissociate fully (since it is a salt). First, find the amount of CH3COO- and H+ dissociated CH3COOH--> CH3COO- + H+ Initial 0.15 0.75 0 Final 0.15-x 0.75 + x x Ka = [CH3COO-][H+]/[CH3CHOOH] (We approximate 0.75 + x to 0.75 since x is a very small number) 1.8 x 10^-5 = [0.75 ] * [x]/[0.15 -x] 2.7 x 10^-6 - 1.8 x 10^-5x = 0.75x (Here, also, approximate 1.8 x 10^-5x to 0) 2.7 x 10^-6 = 0.75 x x= 3.6 x 10^-6 pH = -log(3.6 x 10^-6) = 5.44. Or, to simplify life, you could use pH = pka + log [CH3COO-]/[CH3COOH] BUT, only if the % dissociation is less than 5%. % dissociaion = [conjugate base]/ acid * 100% = [CH3C00-]/[CH3COOH] * 100% Which is in this case, (NOT counting the availability of the 0.75 from the salt), 3.6 x 10^-6 /0.15 = 0.0024%, so the above equation is feasible.
A litre solution containing 0.1M CH3COOH and 0.1M CH3COONa provides a buffer of pH of 4.74. Calculate the pH a?
I suppose that you want the pKa If you have 0.1M of acid and 0.1M of base , you are at half-neutralization and for this point pH=pKa so pKa= 4.74 and you find with the computer Ka=0.000018 the formula for a mixture of salt and acid (buffer ) is pH=pKa+log ([salt]/[acid]) when you add 0.02M of NaOH you neutralize an additional amount of 0.02M so concentration of salt becomes 0.1+0.02=0.12 concentration of acid drops 0.1-0.02=0.08 pH= 4.74+log(1.2/0.8)=4.92
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 !!!
Calculate the final pH of 25 mL of buffer 0.1 M CH3COOH/CH3COONa?
pKa = 4.76 pH = pKa + log(c_base/c_acid) 3.5 = 4.76 + log(c_base/c_acid) - 1.26 = log(c_base/c_acid) 10^-1.26 = c_base/c_acid) 0.055 = c/base/c_acid c_acid + c_base = 0.1 M c_base/c_acid = 0.055 ----- ----> solutions for c_acid and c_base are: c_acid = 0.09447 M c_base = 0.00521 M the added HCl is 0.1/25 M = 0.004 M the new equation is then: pH = 4.76 + log(0.00521 - 0.004)/(0.09447 + 0.004) pH = 4.76 + log(0.00121/0.0985) pH = 4.76 - 1.91 pH = 2.85 OG