Math Snap

STEP 1

What is this asking?
We need to calculate pH, pOH, and concentrations of hydrogen and hydroxide ions given some initial values.
Watch out!
Don't mix up $\text{pH}$ and $\text{pOH}$, and remember those negative exponents!

STEP 2

1. Calculate pH
2. Calculate pOH
3. Calculate hydroxide ion concentration
4. Calculate hydrogen ion concentration

STEP 3

Alright, let's start with calculating the $\text{pH}$!
We're given that the hydrogen ion concentration, $[\text{H}^{+}]$, is $2.7 \cdot 10^{-6}$.

STEP 4

Remember, the formula for $\text{pH}$ is $\text{pH} = -\log_{10}([\text{H}^{+}])$.
Let's plug in our given value:
$$\text{pH} = -\log_{10}(2.7 \cdot 10^{-6})$$

STEP 5

Now, we evaluate the logarithm:
$$\text{pH} \approx -(-5.57)$$

STEP 6

So, our $\text{pH}$ is:
$$\text{pH} \approx 5.57$$

STEP 7

Next up, we're given that the hydroxide ion concentration, $[\text{OH}^{-}]$, is $3.2 \cdot 10^{-8}$.
The formula for $\text{pOH}$ is $\text{pOH} = -\log_{10}([\text{OH}^{-}])$.

STEP 8

Let's plug in our value:
$$\text{pOH} = -\log_{10}(3.2 \cdot 10^{-8})$$

STEP 9

Evaluating the logarithm gives us:
$$\text{pOH} \approx -(-7.49)$$

STEP 10

Therefore, the $\text{pOH}$ is:
$$\text{pOH} \approx 7.49$$

STEP 11

Now, we're given $[\text{H}^{+}] = 5.4 \cdot 10^{-3}$ and we need to find $[\text{OH}^{-}]$.
Remember the key relationship: $[\text{H}^{+}] \cdot [\text{OH}^{-}] = 1.0 \cdot 10^{-14}$.

STEP 12

Let's solve for $[\text{OH}^{-}]$:
$$[\text{OH}^{-}] = \frac{1.0 \cdot 10^{-14}}{[\text{H}^{+}]}$$

STEP 13

Substitute the given value of $[\text{H}^{+}]$:
$$[\text{OH}^{-}] = \frac{1.0 \cdot 10^{-14}}{5.4 \cdot 10^{-3}}$$

STEP 14

Calculating the result:
$$[\text{OH}^{-}] \approx 1.85 \cdot 10^{-12}$$

STEP 15

Finally, we're given $[\text{OH}^{-}] = 1.8 \cdot 10^{-9}$ and we need to find $[\text{H}^{+}]$.
We'll use the same relationship as before: $[\text{H}^{+}] \cdot [\text{OH}^{-}] = 1.0 \cdot 10^{-14}$.

STEP 16

Solving for $[\text{H}^{+}]$:
$$[\text{H}^{+}] = \frac{1.0 \cdot 10^{-14}}{[\text{OH}^{-}]}$$

STEP 17

Substituting the given value of $[\text{OH}^{-}]$:
$$[\text{H}^{+}] = \frac{1.0 \cdot 10^{-14}}{1.8 \cdot 10^{-9}}$$

STEP 18

Calculating the result:
$$[\text{H}^{+}] \approx 5.56 \cdot 10^{-6}$$

(a) $\text{pH} \approx 5.57$
(b) $\text{pOH} \approx 7.49$
(c) $[\text{OH}^{-}] \approx 1.85 \cdot 10^{-12}$
(d) $[\text{H}^{+}] \approx 5.56 \cdot 10^{-6}$