Chapter 8
Chapter 8
Chapter 8
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8.7<br />
Solution<br />
a)<br />
b)<br />
pH 52log 3H 1 4<br />
pH 52log (0.0001)<br />
pH 52(24)<br />
pH 5 4<br />
The pH of the liquid is 4.<br />
pH 52log 3H 1 4<br />
2 52log 3H 1 4<br />
22 5 log 3H 1 4<br />
10 22 5 3H 1 4<br />
0.01 5 3H 1 4<br />
The concentration of hydrogen ions is 0.01 mol>L.<br />
Substitute the value for 3H 1 4<br />
into the equation. Evaluate<br />
log (0.0001).<br />
Substitute the value 2 for the pH.<br />
Divide both sides of the<br />
equation by 21.<br />
Rewrite the equation in<br />
exponential form.<br />
Evaluate the negative<br />
exponent to determine 3H 1 4.<br />
c)<br />
pH 52log 3H 1 4<br />
1.6 52log 3H 1 4<br />
2.5 52log 3H 1 4<br />
To calculate the hydrogen ion<br />
concentration of both<br />
solutions, substitute the given<br />
pH values into the equation.<br />
10 21.6 5 3H 1 4 10 22.5 5 3H 1 4<br />
0.0251 8 3H 1 4 0.0032 8 3H 1 4<br />
0.0251<br />
0.0032 5 7.84<br />
Express both equations in<br />
exponential form, and evaluate.<br />
Divide the concentration of the<br />
first acid by the concentration<br />
of the second acid to find the<br />
relative strength of the acids.<br />
An acid with pH 1.6 is about 7.8 times stronger than an acid with pH 2.5.<br />
NEL <strong>Chapter</strong> 8 495
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