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:?
use the formula;



c1V1 = c2V2



c = concentration (molL^-1)

V = volume (L)

1 = before addition of water

2 = after addition of water



0.50molL^-1 x (20/1000)L = c2 x [(20 + 50)/1000]L

0.01mol = c2 x 0.07L

c2 = 0.01mol / 0.07L

c2 = 0.0007 molL^-1

or 7 x 10^-4 molL^-1



pH = -log[H^+]



pH(before) = -log[0.50] = 0.301029996

pH(after) = -log[7 x 10^-4] = 3.15490196



thus, change in pH =%26gt; pH(after) - pH(before)

= 3.15490196 - 0.301029996

= 2.853871964

thus, the answer is 2.85 (3s.f.)



hop this helps:-)







Hope this helps:-)