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calculate the PH of following solution.
1.40ml 0.05M Na2CO3 + 50 ml of 0.040M of HCL
2.40 ml of 0.020M Na3Po4 + 40 ml of 0.040M of HCL
3. 50 mi of 0.1M Na3PO4 +50ml of 0.1M NaH2PO4
use data the following.
CO2+H20 EQL H+ +HCO3- K1 = 4.2X10^-7
HCO3- EQL H+ + CO3-2 K2=4.8X10^-11
H3PO4 EQL H+ + H2PO4- K1=7.3X10^-3
H2PO4- EQL H+ + HPO4-2 K2=602X10^-8
HPO4-2 EQL H+ + PO4-3 K3=1.0X10^-13
Dear student,
I will give a hint to all the questions:
Calculate the pH of a solution that contains 5.00g of HNO3 in a 2.00 L of a solution?
Sol:
HNO3 is a strong acid. It completely ionizes into H+ and NO3- ions. We first get the molarity of the solution. Before that, we need to convert 5.00 g HNO3 to moles by dividing by its molar mass. 5.00 g / 63.01 g/mol = 0.0794 mol then, we divide 0.0794 by the volume of solution, 2.00 L. M = 0.0794 mol / 2.00 L = 0.0397M Before ionization, the concentration of HNO3 is 0.0397M. After ionization, the concentration of H+ and NO3- ions is also 0.0397M (since the solution completely dissociated, the conjugate base and H+ ions will have equal concentrations). the concentration of H+ ions is also 0.0397M. We now get the pH by getting the negative log of the concentration. pH = -log[H+] = -log (0.0397) = 1.40
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