## 6B 1

rachelle1K
Posts: 109
Joined: Sat Sep 07, 2019 12:16 am

### 6B 1

The question asks for change in pH when HCL concentration is reduced by 12%. I saw a previous chem community post where they plugged in -log(.12) to find the answer, but why don't we plug in H+ (pH= -log([.88])) since it is asking for the value of the change?

Indy Bui 1l
Posts: 99
Joined: Sat Sep 07, 2019 12:19 am

### Re: 6B 1

I believe the question says it is reduced TO 12% of its total, not reduced by 12%.

You would use -Log[0.12] and compared to Log(1), which is 0 the difference is just the value of -Log[0.12]

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

### Re: 6B 1

Doing log[.88] means the concentration was reduced by 12%, but the problem is asking for when it decreases to 12%.

The equation for the initial pH is: $pH = -log[HCl]_i$
Final pH: $pH = -log(0.12[HCl]_i)$
To find the change, just do final pH - initial pH. Through some algebra, you should get $-log(0.12)$ which equals 0.92.

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