Question

a ww.11 mL weak acid that has a concentration of 0.27M will be titrated with 0.24M NaOH. Calculate the pH of the solution upon addition of 0mL of NaOH. The Ka of the acid is 1.6E-6.

a 22.11 mL weak acid that has a concentration of 0.27M will be titrated with 0.24M NaOH. Calculate the pH of the solution upon addition of 0mL of NaOH. The Ka of the acid is 1.6E-6.

Answer 1

Given Molarity of HA = 0.270 M Volume of HA 11.00 ml = 0.011 L We know. Moles = Molarity (M) x Volume (V in L) Moles of HA = 0.270 Mx 0.0110 L = 0.003 mol Molarity of NaOH = 0.240 M 1.Initial pH of solution or at 0.0 ml Base addition Acetic acid is a weak acid, it dissociate partially into acetate and H+ in aqueous media as shown below HA (aq) <-----> A- (aq) + H+ (aq) Initial 0.270 0 -X Equilibrium 0.270 - X Change x) Assume x is small then 0.270 - x will be = 0.270 0.270 Given Ka= 1.60E-06 Weak acid dissociation constant Ka=(A-)(H+)/(HA) 1.60E-06 = x*x+( 0.270 - 1.60E-06 = x*x+ 0.27 x^2 1.60E-06 $ sox^2 4.32E-07 X = 0.00066 but from ICE table x = [H+); So [H+] = 0.0006573 but pH = -log [H+] = -log[ 0.00065727 ] = 3.18226 so pH of solution = M

Given Molarity of HA = 0.270 M Volume of HA 22.10 ml = 0.0221 L We know. Moles = Molarity (M) x Volume (V in L) Moles of HA = 0.270 Mx 0.0221 L = 0.006 mol Molarity of NaOH = 0.240 M 1.Initial pH of solution or at 0.0 ml Base addition Acetic acid is a weak acid, it dissociate partially into acetate and H+ in aqueous media as shown below HA (aq) <-----> A- (aq) + H+ (aq) Initial 0.270 0 -X Equilibrium 0.270 - X Change + X x) Assume x is small then 0.270 - x will be = 0.270 0.270 Given Ka= 1.60E-06 Weak acid dissociation constant Ka=(A-)(H+)/(HA) 1.60E-06 = x*x+( 0.270 - 1.60E-06 = x*x+ 0.27 x^2 1.60E-06 sox^2 4.32E-07 X = 0.00066 but from ICE table x = [H+]; So [H+] = 0.0006573 but pH = -log [H+] = -log[ 0.00065727 ] = 3.18226 so pH of solution = 3.18 = M

note : PH is same for weak acid weather it is 11 ml or 22 ml for zero volume base addition