Math / Numbers & Operations

Question(a) The $\left[\mathrm{H}^{+}\right]$ of a solution is $2.7 \times 10^{-6}$ Calculate the pH . (b) The $\left[\mathrm{OH}^{-}\right]$ of a solution is $3.2 \times 10^{-8}$ Calculate the pOH . (c) The $\left[\mathrm{H}^{+}\right]$ of a solution is $5.4 \times 10^{-3}$ Calculate the $\left[\mathrm{OH}^{-}\right]$ . (d) The $\left[\mathrm{OH}^{-}\right]$ of a solution is $1.8 \times 10^{-9}$ Calculate the $\left[\mathrm{H}^{+}\right]$ .

Studdy Solution

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

STEP 19

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

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Studdy Solution

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 pH\text{pH}pH and pOH\text{pOH}pOH, and remember those negative exponents!

STEP 2

1. Calculate pH2. Calculate pOH3. Calculate hydroxide ion concentration4. Calculate hydrogen ion concentration

STEP 3

Alright, let's **start** with calculating the pH\text{pH}pH!We're given that the hydrogen ion concentration, [H+][\text{H}^{+}][H+], is 2.7⋅10−62.7 \cdot 10^{-6}2.7⋅10−6.

STEP 4

Remember, the formula for pH\text{pH}pH is pH=−log⁡10([H+])\text{pH} = -\log_{10}([\text{H}^{+}])pH=−log10​([H+]).Let's plug in our **given value**:pH=−log⁡10(2.7⋅10−6) \text{pH} = -\log_{10}(2.7 \cdot 10^{-6}) pH=−log10​(2.7⋅10−6)

STEP 5

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

STEP 6

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

STEP 7

Next up, we're given that the hydroxide ion concentration, [OH−][\text{OH}^{-}][OH−], is 3.2⋅10−83.2 \cdot 10^{-8}3.2⋅10−8.The formula for pOH\text{pOH}pOH is pOH=−log⁡10([OH−])\text{pOH} = -\log_{10}([\text{OH}^{-}])pOH=−log10​([OH−]).

STEP 8

Let's **plug in** our value:pOH=−log⁡10(3.2⋅10−8) \text{pOH} = -\log_{10}(3.2 \cdot 10^{-8}) pOH=−log10​(3.2⋅10−8)

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

**Substitute** the given value of [H+][\text{H}^{+}][H+]:[OH−]=1.0⋅10−145.4⋅10−3 [\text{OH}^{-}] = \frac{1.0 \cdot 10^{-14}}{5.4 \cdot 10^{-3}} [OH−]=5.4⋅10−31.0⋅10−14​

STEP 14

**Calculating** the result:[OH−]≈1.85⋅10−12 [\text{OH}^{-}] \approx 1.85 \cdot 10^{-12} [OH−]≈1.85⋅10−12

STEP 15

STEP 16

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

STEP 17

**Substituting** the given value of [OH−][\text{OH}^{-}][OH−]:[H+]=1.0⋅10−141.8⋅10−9 [\text{H}^{+}] = \frac{1.0 \cdot 10^{-14}}{1.8 \cdot 10^{-9}} [H+]=1.8⋅10−91.0⋅10−14​

STEP 18

**Calculating** the result:[H+]≈5.56⋅10−6 [\text{H}^{+}] \approx 5.56 \cdot 10^{-6} [H+]≈5.56⋅10−6

STEP 19

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

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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!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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) $[\text{H}^{+}] \approx 5.56 \cdot 10^{-6}$