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Question: Urine normally has a pH of 6. If a patient eliminates 1.3 liters of urine per day, how many moles of...

Urine normally has a pH of 6. If a patient eliminates 1.3 liters of urine per day, how many moles of H+{H^ + } ions does he urinate?
A. 1.3×10−31.3 \times {10^{ - 3}}
B. 1.3×10−61.3 \times {10^{ - 6}}
C. 2.4×10−72.4 \times {10^{ - 7}}
D. 2.4×10−62.4 \times {10^{ - 6}}

Explanation

Solution

The term pH denotes the ‘potential of hydrogen’ or ‘power of hydrogen’ and is used to measure the quantitative acidity or basicity of an aqueous or other liquid solution. The pH scale is logarithmic in nature

Complete step by step answer:
As we know the pH for any fluid is the measure of its hydrogen ion concentration relative to that of a given standard solution. Mathematically pH for any liquid is given as the logarithm(to the base 10) of the reciprocal of the hydrogen ion concentration. Or we can write:
pH=−log⁡[H+]pH = - \log [{H^ + }]
Here given that the pH of urine is 6. Hence from the above equation we get:
6=−log⁡[H+]6 = - \log [{H^ + }]
Or
[H+]=10−6[{H^ + }] = {10^{ - 6}}
Hence the molar concentration of hydrogen ion is 10−6mol/lt{10^{ - 6}}mol/lt. That means one liter of urine contains = 10−6mol{10^{ - 6}}molof urine.
Then 1.3 liter of urine contains = 1.3×10−6mol1.3 \times {10^{ - 6}}mol of urine

So, the correct answer is Option B.

Note: The pH of any solution is generally defined on the scale from 0-14 with lesser number indicating more acidic behaviour. Pure water is considered to be neutral and has a pH of 7. Also the pH of any solution is inversely proportional to the concentration of H+{H^ + } ions and directly proportional to the concentration of OH−O{H^ - } ions.
With each whole number change in pH there is 10 times change in concentration of hydrogen ions. Having knowledge of the pH of any solution is highly useful and very important as it represents the chemical conditions of a solution. Also many minerals can become hazardous in acidic environments that may be neutral in a basic environment.