6B.1

[Chris Tai 1B]

The molar concentration of HCl in hydrochloric acid is reduced to 12% of its initial value by dilution. What is the difference in the pH values of the two solutions?

What equations would I use to set up this problem?

[nicolely2F]

The pH equation is -log([H+]). The equation for change is -log(change in concentration / original concentration). In this case: -log(0.12*[HCl] / [HCl]). We don't need the actual value of [HCl], we can just assume it's 1 to represent the original value. Thus, we have -log(0.12) = 0.92

[Tiffany Vo 3G]

I don't know if this is easier, but I made the molarity of the original solution equal to 100M and the second solution's molarity equal to 12% of the original, or 12 M. It'd allow you to have actual numbers for the H3O+ concentration at least. You could then find out the pH of each solution and subtract the second solution's pH from the first one's pH to find the difference.

[rachelle1K]

The pH equation is -log([H+]). The equation for change is -log(change in concentration / original concentration). In this case: -log(0.12*[HCl] / [HCl]). We don't need the actual value of [HCl], we can just assume it's 1 to represent the original value. Thus, we have -log(0.12) = 0.92

Why don't we plug in H+ (pH= -log([.88])) since it is asking for the value of ?