## Test 2

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Test 2

5. The ionic dissociation of water is given by the following reaction: The ΔH for the reaction is 58kJ/mol. The Kw for the reaction at 25°C is 10^-14. Is a pH of 7 acidic or basic at 30°C?
2H2O(l) $\leftrightarrow$ H30+(aq) + OH-(aq)

What is the pH? I used the nerst equation but I'm not getting the right answer.

armintaheri
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

### Re: Test 2

You have to use the Van't Hoff equation:

$ln(\frac{k_{2}}{k_{1}})=-\frac{\Delta H}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$

Plug in values:

$ln(\frac{k_{2}}{10^{-14}})=-\frac{58000}{8.314}(\frac{1}{303.15}-\frac{1}{298.15})$

Solve for $k_{2}$:

$k_{2}=1.3\times 10^{-14}$

$pk_{2}=-log(1.3\times 10^{-14}) = 13.90$

Since for a neutral solution $pH=pOH$:

$pk_{2}= pH+pOH = pH + pH = 2pH$

$pH=\frac{pk_{2}}{2}=\frac{13.90}{2}=6.95$

Therefore, a neutral pH is 6.95 at 30°C. Since 7 is greater than 6.95, a pH of 7 is basic at 30°C.

Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Re: Test 2

How is there 2pH?

Faith Fredlund 1H
Posts: 68
Joined: Fri Sep 28, 2018 12:18 am

### Re: Test 2

The above work shows 2pH because they are showing that in a neutral solution, pH=7 and pOH=7, so because they have the same value, you can take the equation

pH + pOH = 14

and substitute pH for the pOH and get,

pH + pH = 14 or 2pH = 14

Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Re: Test 2

So can we assume we always assume that when the pH is 7 then it will be pH + pH = 14 in which we will divide by 2.

Lynsea_Southwick_2K
Posts: 55
Joined: Fri Sep 28, 2018 12:25 am

### Re: Test 2

How do you know that it is a neutral solution?

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