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Calculate the hydrogen ion concentration in the following biological fluids whose \(\mathrm{pH}\) are given below:
(a) Human muscle-fluid, \(6.83\)
(b) Human stomach fluid, \(1.2\)
(c) Human blood, \(7.38\)
(d) Human saliva, 6.4
(a) Human muscle-fluid, \(6.83\)
(b) Human stomach fluid, \(1.2\)
(c) Human blood, \(7.38\)
(d) Human saliva, 6.4
Solution:
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Verified Answer
(a) \(\log \left[\mathrm{H}^{+}\right]=\mathrm{pH}=-6.83\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-6.83=1.479 \times 10^{-7} \mathrm{M}\)
(b) \(\log \left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-1.2\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-1.2=6.31 \times 10^{-2} \mathrm{M}\)
(c) \(\log \left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-7.38\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-7.38=4.169 \times 10^{-8} \mathrm{M}\)
(d) \(\log \left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-6.4\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-6.4=3.981 \times 10^{-7} \mathrm{M}\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-6.83=1.479 \times 10^{-7} \mathrm{M}\)
(b) \(\log \left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-1.2\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-1.2=6.31 \times 10^{-2} \mathrm{M}\)
(c) \(\log \left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-7.38\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-7.38=4.169 \times 10^{-8} \mathrm{M}\)
(d) \(\log \left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-6.4\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(-6.4=3.981 \times 10^{-7} \mathrm{M}\)
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