Calculate the change in PH that occurs?

I have already posted this question but am a little confused since I received 3 different responses with 3 different answers. I am not sure which of the responses is correct, so I ask someone who is very confident of their response to please explain how to do this problem. I know that CHANGE in ph is to the power of 10 so perhaps this makes a difference in the actual response. Not sure but thanks again!





Calculate the change in PH that occurs:



50 ml of water is added to 20 ml of a 0.50 mol/L solution of hydrochloric acid. Calculate the CHANGE in ph that occurs.



Please show the steps how to do this problem.Calculate the change in PH that occurs?
*****step 1.....



pH 鈮?-log {H+} where {H+} is in moles / liter



鈭?pH = (-log {diluted HCl}) - (-log {.50 M HCl})





*****step 2. find concentration of diluted HCl...



can be found from....



M1V1 = M2V2

M2 = M1V1/V2 = .50 M x (20 ml ) / ( 20 + 50 ml) = .143 M



note that V2 = 20 ml + 50 ml (you started with 20 and added 50.....)





****step 3 calculate 鈭?pH



so 鈭?pH = (-log {.143}) - (-log {.50})= (.845) - (.301)= .544



should be reported as an increase by 0.54 pH units since you only have 2 sig figs.....Calculate the change in PH that occurs?
PH is defined as the negative log of the hydrogen ion concentration. This concentration is expressed in terms of moles/liter. A 0.5 mol/l solution of HCl contains half a mole of HCl per liter. There are 1000 ml in a liter, so 20 ml of this solution would contain (20/1000)*0.5 moles/liter, or 0.01 moles. Now, the concentration changes because an additional 50 ml of water is added. The total volume is now 70ml, but this still contains 0.01 moles. The new concentration is now 0.01 moles / (70/1000) liters, or 0.143 moles/liter.



The initial concentration was 0.5 moles/liter. The negative log of 0.5 is .3010 and this is the initial pH. After the water is added, the concentration is 0.143, and the negative log of this is 0.845, so the change in pH is the difference, or 0.544 and in terms of pH, the solution has become more alkaline.