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pankaj kalal Grade:
        


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


6 years ago

Answers : (2)

Vijay Luxmi Askiitiansexpert
357 Points
										

Dear Pankaj,


 


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

6 years ago
SAGAR SINGH - IIT DELHI
879 Points
										

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.


 



























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6 years ago
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