6B.1

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Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

6B.1

Postby Chris Tai 1B » Tue Dec 03, 2019 5:40 pm

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
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: 6B.1

Postby nicolely2F » Wed Dec 04, 2019 2:31 am

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
Posts: 52
Joined: Sat Aug 17, 2019 12:17 am

Re: 6B.1

Postby Tiffany Vo 3G » Wed Dec 04, 2019 10:20 am

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
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

Re: 6B.1

Postby rachelle1K » Wed Dec 04, 2019 4:24 pm

nicolely3B wrote: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 ?


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