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Exercises · 6.55

Q.Calculate the hydrogen ion concentration in the following biological fluids whose 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.
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The hydrogen ion concentration is found using [H+]=10−pH[H^+] = 10^{-\text{pH}}. For muscle fluid (pH 6.83) it is 1.48×10−7 M1.48 \times 10^{-7} \text{ M}, for stomach fluid (pH 1.2) it is 6.31×10−2 M6.31 \times 10^{-2} \text{ M}, for blood (pH 7.38) it is 4.17×10−8 M4.17 \times 10^{-8} \text{ M}, and for saliva (pH 6.4) it is 3.98×10−7 M3.98 \times 10^{-7} \text{ M}.

The pH scale is a logarithmic measure of acidity. The definition is simple but powerful:

pH=−log⁡10[H+]\text{pH} = -\log_{10}[H^+]

Here [H+][H^+] is the molar concentration of hydrogen ions (in mol/L). The negative sign means that a lower pH corresponds to a higher hydrogen ion concentration — that’s why stomach acid (pH ~1.2) has far more H+H^+ than blood (pH ~7.38).

To go backwards — from pH to [H+][H^+] — we invert the logarithm. If pH=−log⁡[H+]\text{pH} = -\log[H^+], then multiplying both sides by -1 gives log⁡[H+]=−pH\log[H^+] = -\text{pH}. Raising 10 to both sides:

[H+]=10−pH[H^+] = 10^{-\text{pH}}

That’s the only formula you need for all four parts. Each is a direct substitution.

1. Human muscle fluid, pH = 6.83

[H+]=10−6.83[H^+] = 10^{-6.83}

How do you evaluate 10−6.8310^{-6.83} without a calculator in an exam? Write it as:

10−6.83=10−7×100.1710^{-6.83} = 10^{-7} \times 10^{0.17}

Now 100.1710^{0.17} is the antilog of 0.17. You can recall that log⁡1.48≈0.17\log 1.48 \approx 0.17 (since log⁡1.5≈0.176\log 1.5 \approx 0.176). So 100.17≈1.4810^{0.17} \approx 1.48.

Thus:

[H+]≈1.48×10−7 M[H^+] \approx 1.48 \times 10^{-7} \text{ M}

Tip

For quick antilog: if the decimal part is xx, find the number whose log is xx. Common values: log⁡1.0=0\log 1.0 = 0, log⁡1.2≈0.079\log 1.2 \approx 0.079, log⁡1.5≈0.176\log 1.5 \approx 0.176, log⁡2.0≈0.301\log 2.0 \approx 0.301. So 100.1710^{0.17} lies between 1.48 and 1.5.

2. Human stomach fluid, pH = 1.2

[H+]=10−1.2=10−2×100.8[H^+] = 10^{-1.2} = 10^{-2} \times 10^{0.8}

Now 100.810^{0.8}: log⁡6.31≈0.8\log 6.31 \approx 0.8 (since log⁡6.3≈0.799\log 6.3 \approx 0.799). So:

[H+]≈6.31×10−2 M[H^+] \approx 6.31 \times 10^{-2} \text{ M}

That’s about 0.063 M — a strongly acidic solution, as expected for gastric juice. …

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