Calculate the pH value in each of the following solutions, given their [H3O+] concentrations. Solution [H3O+](M)
gastric juice 1.6x10^-2
cow's milk 2.5x10^-7
tomato juice 5.0x10^-5

See Explanations ...

Explanation:

In general, pH is a 'p-factor' expression which, simply put, is a way to express very small numbers (i.e.; exponential data with 10⁻ⁿ value ranges) in a more convenient form. That is, by definition, pX = -log(X) where X is the data value of interest. In practical terms, p-factor analysis can be applied to a number of physical & chemical measurements such as ...

pH => measure of acidity of solution = -log[H₃O⁺]

pOH => measure of alkalinity of solution = -log[OH⁻]

pKa => measure of weak acid ionization in aqueous solution = -log(Ka)

pKb => measure of weak base ionization in aqueous solution = -log(Kb)

pKsp => measure of salt ionization in aqueous solution = -log(Ksp)

Such can be applied to ranges of small-number values defining other chemical and physical properties.

For this problem:

Gastric Juice: [H₃O⁺] = 1.6 x 10⁻²M => pH = -log(1.6 x 10⁻²) = -(-1.80) = 1.8

Cow's Milk:  [H₃O⁺] = 2.5 x 10⁻⁷M => pH = -log(2.5 x 10⁻⁷) = -(-6.60) = 6.60

Tomato Juice:  [H₃O⁺] = 5.0 x 10⁻⁵M => pH = -log(5.0 x 10⁻⁵) = -(-4.30) = 4.30

Other Applications:

Given:

[OH⁻] = 6.30 x 10⁻¹³M => pOH = -log(6.30 x 10⁻¹³) = -(-12.2) = 12.2

Ka = 4.5 x 10⁻⁵ => pKa = -log(4.5 x 10⁻⁵) = -(-4.35) = 4.35

Kb = 8.2 x 10⁻⁶ => pKb = -log(8.2 x 10⁻⁶) = -(-5.09) = 5.09

Ksp = 5.5 x 10⁻¹⁰ => pKsp = -log(-5.5 x 10⁻¹⁰) = -(-9.26) = 9.26

Note: The values for Ka, Kb & Ksp are typically provided in tables of weak acid ionization constants (Ka-values), weak base ionization constants (Kb-values) or solubility product constants of salts (Ksp-values).

Hope this helps, Doc :-)

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2