Question

if the poh of vinegar is 9.45, what is its \\(\text{oh}^-\\)?
round to the nearest hundredth.

\\(\square \times 10^n\text{ m}\\) \\(n = \square\\)

Explanation:

Identify the given values and formula

We are given the \(\text{pOH}\) of vinegar:

$$\text{pOH} = 9.45$$

We need to find the hydroxide ion concentration, \([\text{OH}^-]\), expressed in scientific notation as:

$$\text{coefficient} \times 10^n\text{ M}$$

The relationship between \([\text{OH}^-]\) and \(\text{pOH}\) is given by the formula:

$$[\text{OH}^-] = 10^{-\text{pOH}}$$

Calculate the hydroxide concentration

Substitute the given \(\text{pOH}\) value into the formula:

$$[\text{OH}^-] = 10^{-9.45}$$

Convert to standard scientific notation

To express \(10^{-9.45}\) in standard scientific notation \(a \times 10^n\), where \(1 \le a < 10\) and \(n\) is an integer:

$$10^{-9.45} = 10^{0.55 - 10} = 10^{0.55} \times 10^{-10}$$

Now, calculate the value of \(10^{0.55}\):

$$10^{0.55} \approx 3.548$$

Round to the nearest hundredth

Rounding the coefficient \(3.548\) to the nearest hundredth gives:

$$3.55$$

Thus, the concentration is:

$$3.55 \times 10^{-10}\text{ M}$$

Comparing this to the template \(\text{[blank]} \times 10^n\text{ M}\) where \(n = \text{[blank]}\):

• The coefficient blank is \(3.55\)
• The exponent \(n\) is \(-10\)

Answer:

If the pOH of vinegar is 9.45, what is its \([\text{OH}^-]\)? Round to the nearest hundredth.

<blank>\(3.55\)</blank> \(\times 10^n\text{ M}\) \(n =\) <blank>\(-10\)</blank>