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 Pankaj,

You have Posted the same question Twice... Please post the question once...

Dear student,

Use this hint in these questions:

When the [H₃O⁺] is 10^-6 M, the pH is 6. Also, the [OH^-] is 10^-8 M and the pOH is 8. Again, the pH and the pOH add up to 14. When the [H₃O⁺] is 10^-5 M, pH is 5, [OH^-] is 10^-9 M, and pOH is 9. pH + pOH = 14.

However, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. For example, we skipped over the value of 2.0 x 10^-7. This is more complicated. However, if you use a calculator that will handle logarithms, it is a very simple calculation. First you enter the hydronium ion concentration. You can use decimal format or scientific notation. Next push the log button. Then change the sign by pushing the +/- button. In this case we get 6.70 for the pH. The other values can be obtained in the same way.

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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