Question

the concentration of hydronium ions and hydroxide ions can vary widely in a solution, easily from 1 m to 1 x 10⁻¹⁴ m. this variation is too wide to fit on an analog readout. in order to make this into a smaller scale suitable for an analog readout, the log to the base 10 was used so that the read out is focused on powers of 10. note that the log 1 = 0, and the log of 1 x 10⁻¹⁴ = -14. it was then decided that the numbers should be positive so the sign was changed. the way to measure concentrations of both h⁺ and oh⁻ was thus defined as -log. this was then referred to as p. thus, ph = -logh⁺, poh = -logoh⁻, and pk = -logkₑq. you need to figure out how to do this mathematical manipulation on your calculator. the acid - base reaction in pure water is h₂o + h₂o ⇌ h₃o⁺¹ + oh⁻¹ and it was found that in pure water under standard conditions the h₃o⁺¹ = oh⁻¹ = 1 x 10⁻⁷ m neutral ph is defined as having a h⁺ or h₃o⁺¹ of 1 x 10⁻⁷ m. the ph of a neutral solution is thus ph = -log1 x 10⁻⁷ = 1 the poh of a neutral solution is thus poh = -log1 x 10⁻⁷ = 2 a. 7 b. 6 c. 5. d. 4 e. 3 f. 2 g. 1

Explanation:

Step1: Recall log properties

We know that \(\log(1\times10^{n}) = n\) (since \(\log(a\times10^{b})=\log(a) + b\) and \(\log(1) = 0\)). For \(\log(1\times10^{-7})\), this gives \(\log(1\times10^{-7})=-7\).

Step2: Apply the -log operation for pH (and pOH)

The formula for pH (and pOH in this case) is \(-\log[\text{ion concentration}]\). For \([\text{H}_3\text{O}^+]=1\times 10^{-7}\space M\) (or \([\text{OH}^-]=1\times 10^{-7}\space M\) for pOH), we calculate \(-\log(1\times 10^{-7})\). Substituting the value from Step 1, we get \(-(-7)=7\).

Answer:

For both the pH (blank 1) and pOH (blank 2) of a neutral solution, the answer is A. 7.