Calculate the pH of 1.00 x 10^-2 M oxalic acid solution. Calculate the pH of 1 x 10^-2 M solution of sodium oxalate. Ka1 = 5.62 x 10^-2, Ka2 = 5.42 x 10^-5
For the vast majority of polyprotic acid pH questions, onemust only consider the disassociation of the first proton, thesecond/third/etc protons are much harder to lose and thus theconcentration of H+ from those disassociations are negligible forall practical purposes.
First, we do an ICE table:
1 x 10^-2-x +x +x
Ka1=5.62e-2=(x^2)/( 1 x 10^-2-x)
x^2+5.62e-2*x-5.62e-4=0
Then, solve this quadratic equation, we get x = [H+]= 0.065M
pH = -log[H+] = 1.19
Calculate the pH of 1.00 x 10^-2 M oxalic acid solution. Calculate the pH of 1...
calculate the pH and pOH for a 0.15 M oxalic acid H2C2O4 solution. (ka1-6.0x10^-2, Ka2-6.1x10^-5)
A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH. Calculate the pHs when 25.00 mL of oxalic acid solution is titrated with 10, 15, 20, 25, 35, 40, 45, 50 and 55 mL of NaOH. Kal = 5.4 x 10-2 and Ka2 = 5.42 x 10-5 for oxalic acid. Show all your work and make a graph of pH versus Vb
A 27 mL sample of 0.100 M oxalic acid (Ka1= 5.60 * 10^-2, Ka2= 5.42* 10^-5) was titrated with 0.0850 M NaOH. What is the pH at 4 mL before the equivalence point? What is the pH at 4 mL after the equivalence point?
Calculate the pH and the equilibrium concentrations of HC2O4- and C2O42- in a 0.0782 M oxalic acid solution, H2C2O4 (aq). For H2C2O4, Ka1 = 5.9×10-2 and Ka2 = 6.4×10-5 pH =_______ [HC2O4-] =___________ M [C2O42-] =___________ M
Calculate the pH and the equilibrium concentrations of HC2O4- and C2O42- in a 0.1160 M oxalic acid solution, H2C2O4 (aq). For H2C2O4, Ka1 = 5.9×10-2 and Ka2 = 6.4×10-5 pH = [HC2O4-] = M [C2O42-] = M
A 27 mL sample of 0.100 M oxalic acid (Ka1= 5.60 * 10^-2, Ka2= 5.42* 10^-5) was titrated with 0.0850 M NaOH. What is the pH at 4 mL before the equivalence point? What is the pH at 4 mL after the equivalence point?
A titration is carried out for 20.0mL of 0.10 M Oxalic Acid (weak acid) with 0.10 M of a strong base NaOH. Calculate the pH at these volumes of added base solution: (a) 0.0 mL (b) 5.0 mL (c) 10.0 mL (d) 15.0 mL (e) 20.0 mL (f) 25.0 mL (g) 30.0 mL Oxalic acid Ka1 = 5.9 x 10-2 Ka2 = 6.4 x 10-5
What is the hydroxide-ion concentration of a 0.250 M sodium oxalate (Na2C2O4) solution?For oxalic acid (H2C2O4), Ka1 = 5.6×10–2 and Ka2 = 5.1×10–5.
Calculate the pH of each solution. A solution containing 0.0320 M maleic acid and 0.046 M disodium maleate. The Ka values for maleic acid are 1.20 x 10 (Ka1) and 5.37 x 10-7 (Ka2) 1.8 pH A solution containing 0.0306 M succinic acid and 0.019 M potassium hydrogen succinate. The Ka values for succinic acid are 6.21 x 105 (Kal) and 2.31 x 10 (Ka2) pH
Calculate the pH of each solution. A solution containing 0.0345 M0.0345 M maleic acid and 0.047 M0.047 M disodium maleate. The ?aKa values for maleic acid are 1.20×10−2 (?a1)1.20×10−2 (Ka1) and 5.37×10−7 (?a2).5.37×10−7 (Ka2). pH= A solution containing 0.0320 M0.0320 M succinic acid and 0.018 M0.018 M potassium hydrogen succinate. The ?aKa values for succinic acid are 6.21×10−5 (?a1)6.21×10−5 (Ka1) and 2.31×10−6 (?a2).2.31×10−6 (Ka2). pH=