For 0.045 M H2CO3, calculate [H30+],[HCO3],[CO3^-2], given Ka1= 4.4 x 10^-7 and Ka2= 4.7 x 10^-11.
For 0.045 M H2CO3, calculate [H30+],[HCO3],[CO3^-2], given Ka1= 4.4 x 10^-7 and Ka2= 4.7 x 10^-11.
Calculate the pH and concentration of CO3^2- in a 0.25 M solution of carbonic acid, H2CO3. Ka1= 4.4 x 10^-7 Ka2= 4.7 x 10^-11
Calculate the pH and the concentrations of all species present (H2CO3, HCO3 – , CO3 2– , H3O + , and OH– ) in a 0.0037 M M carbonic acid solution. Ka1 = 4.3 × 10–7 ; Ka2 = 5.6 × 10–11
The following questions is based on ionization of carbonic acid. [H2CO3 = H+ + HCO3- K1 = 4.7 X 10-7, pK1 = 6.34 ] [HCO3- = H+ CO3-2 K2 = 4.4 X 10-11, pK2 = 10.36] Calculate the pH ofthe solution form by mixing the following: a)50.0ml of 0.100 M H2CO3 with 50.0ml of 0.050 M NaOH . b)50.0ml of 0.100 M H2CO3 with 50.0ml of 0.150 M NaOH . c)50.0ml of 0.100 M H2CO3 with 50.0ml of 0.100 M...
Calculate the pH of 0.0150 M H2CO3 For H2CO3, Ka1 = 4.46 x 10-7, Ka2 = 4.69 x 10-11 Answer: pH = 4.09 (acidic form for diprotic) Please help me show work! I don't understand how to solve problems like this. The answer is here.
What is the pH of 0.10 M NaHCO3? Ka1 of H2CO3: 4.3E-7 Ka2 of H2CO3: 5.6E-11
Carbonic acid, H2CO3 is a diprotic acid with Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11. What is the pH of a 0.47 M solution of carbonic acid?
Calculate the concentration of all species in a 0.165 M solution of H2CO3 (Ka1 is 4.3×10−7 and Ka2 is 5.6×10−11).
Calculate the pH of 0.020M carbonic acid, H2CO3 after both dissociations. Ka1=4.45x10^-7, Ka2=4.69x10^-11
You create a 1 L solution of 0.1 M H2CO3. carbonic acid, H2CO3, is a diprotic acid with Ka1 = 4.5 x 10-7 and Ka2 = 4.7 x 10-11. a) What will the initial pH of the solution be? b) What volume of 0.1 M NaOH will you need to add to reach the second equivalence point( remember carbonic acid deprotonates to bicarbonate HCO3- and then can deprotonate further to CO32-? c) At the second equivalence point, what will the...
What is the pH of a 0.10 M solution of carbonic acid? Carbonic acid, H2CO3 has two acidic protons: H2CO3 + H2O7 HCO3 + H30+ Ka1 = 4.3x10-7 HCO3 + H202 CO32- + H30+ Ka2 = 5.6x10-11 a) 1.00 b) 0.70 c) 6.37 d) 3.68 e) I still can't figure this out...