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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
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 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.
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