calculate the PH of following solution.1.40ml 0.05M Na2CO3 + 50 ml of 0.040M of HCL2.40 ml of 0.020M Na3Po4 + 40 ml of 0.040M of HCL3. 50 mi of 0.1M Na3PO4 +50ml of 0.1M NaH2PO4use data the following.CO2+H20 EQL H+ +HCO3- K1 = 4.2X10^-7HCO3- EQL H+ + CO3-2 K2=4.8X10^-11H3PO4 EQL H+ + H2PO4- K1=7.3X10^-3H2PO4- EQL H+ + HPO4-2 K2=602X10^-8HPO4-2 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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