QuestionFinding the conjugate of an acid or base
Fill in the missing chemical formulae in the tables below. \begin{tabular}{|c|c|} \hline acid & conjugate base \\ \hline $\mathrm{NH}_{4}^{+}$ & $\square$ \\ \hline $\mathrm{H}_{2} \mathrm{CO}_{3}$ & $\square$ \\ \hline HI & $\square$ \\ \hline \end{tabular} \begin{tabular}{|c|c|} \hline base & conjugate acid \\ \hline $\mathrm{HPO}_{4}^{2-}$ & $\square$ \\ \hline $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ & $\square$ \\ \hline $\mathrm{NH}_{3}$ & $\square$ \\ \hline \end{tabular}

Studdy Solution

STEP 1

What is this asking? We need to find the conjugate bases of some acids and the conjugate acids of some bases. Watch out! Remember, a conjugate base is formed when an acid loses a proton ( $H^{+}$ ), and a conjugate acid is formed when a base gains a proton ( $H^{+}$ ).
Don't accidentally add or subtract other atoms!

STEP 2

1. Conjugate Bases
2. Conjugate Acids

STEP 3

Let's start with $\mathrm{NH}_{4}^{+}$ .
To find its conjugate base, we remove one $H^{+}$ .
So, $\mathrm{NH}_{4}^{+} - H^{+} = \mathrm{NH}_{3}$ .
The conjugate base of $\mathrm{NH}_{4}^{+}$ is $\mathrm{NH}_{3}$ !

STEP 4

Next up is $\mathrm{H}_{2}\mathrm{CO}_{3}$ . Removing one $H^{+}$ gives us $\mathrm{H}_{2}\mathrm{CO}_{3} - H^{+} = \mathrm{HCO}_{3}^{-}$ .
The conjugate base of $\mathrm{H}_{2}\mathrm{CO}_{3}$ is $\mathrm{HCO}_{3}^{-}$ !

STEP 5

Finally, we have HI. Subtracting an $H^{+}$ gives us $HI - H^{+} = I^{-}$ .
The conjugate base of HI is $I^{-}$ !

STEP 6

Now, let's find the conjugate acids.
We'll begin with $\mathrm{HPO}_{4}^{2-}$ .
To find its conjugate acid, we add one $H^{+}$ .
So, $\mathrm{HPO}_{4}^{2-} + H^{+} = \mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ .
The conjugate acid of $\mathrm{HPO}_{4}^{2-}$ is $\mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ !

STEP 7

Next, we have $\mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ . Adding one $H^{+}$ gives us $\mathrm{H}_{2}\mathrm{PO}_{4}^{-} + H^{+} = \mathrm{H}_{3}\mathrm{PO}_{4}$ .
The conjugate acid of $\mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ is $\mathrm{H}_{3}\mathrm{PO}_{4}$ !

STEP 8

Lastly, we have $\mathrm{NH}_{3}$ . Adding an $H^{+}$ gives us $\mathrm{NH}_{3} + H^{+} = \mathrm{NH}_{4}^{+}$ .
The conjugate acid of $\mathrm{NH}_{3}$ is $\mathrm{NH}_{4}^{+}$ !

STEP 9

The completed table should look like this:
Acid | Conjugate Base ---|--- $\mathrm{NH}_{4}^{+}$ | $\mathrm{NH}_{3}$ $\mathrm{H}_{2} \mathrm{CO}_{3}$ | $\mathrm{HCO}_{3}^{-}$ HI | $I^{-}$
Base | Conjugate Acid ---|--- $\mathrm{HPO}_{4}^{2-}$ | $\mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}$ | $\mathrm{H}_{3}\mathrm{PO}_{4}$ $\mathrm{NH}_{3}$ | $\mathrm{NH}_{4}^{+}$

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Studdy Solution

STEP 1

STEP 2

1. Conjugate Bases
2. Conjugate Acids

STEP 3

Let's start with $\mathrm{NH}_{4}^{+}$ .
To find its conjugate base, we remove one $H^{+}$ .
So, $\mathrm{NH}_{4}^{+} - H^{+} = \mathrm{NH}_{3}$ .
The conjugate base of $\mathrm{NH}_{4}^{+}$ is $\mathrm{NH}_{3}$ !

STEP 4

Next up is $\mathrm{H}_{2}\mathrm{CO}_{3}$ . Removing one $H^{+}$ gives us $\mathrm{H}_{2}\mathrm{CO}_{3} - H^{+} = \mathrm{HCO}_{3}^{-}$ .
The conjugate base of $\mathrm{H}_{2}\mathrm{CO}_{3}$ is $\mathrm{HCO}_{3}^{-}$ !

STEP 5

Finally, we have HI. Subtracting an $H^{+}$ gives us $HI - H^{+} = I^{-}$ .
The conjugate base of HI is $I^{-}$ !

STEP 6

Now, let's find the conjugate acids.
We'll begin with $\mathrm{HPO}_{4}^{2-}$ .
To find its conjugate acid, we add one $H^{+}$ .
So, $\mathrm{HPO}_{4}^{2-} + H^{+} = \mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ .
The conjugate acid of $\mathrm{HPO}_{4}^{2-}$ is $\mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ !

STEP 7

Next, we have $\mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ . Adding one $H^{+}$ gives us $\mathrm{H}_{2}\mathrm{PO}_{4}^{-} + H^{+} = \mathrm{H}_{3}\mathrm{PO}_{4}$ .
The conjugate acid of $\mathrm{H}_{2}\mathrm{PO}_{4}^{-}$ is $\mathrm{H}_{3}\mathrm{PO}_{4}$ !

STEP 8

Lastly, we have $\mathrm{NH}_{3}$ . Adding an $H^{+}$ gives us $\mathrm{NH}_{3} + H^{+} = \mathrm{NH}_{4}^{+}$ .
The conjugate acid of $\mathrm{NH}_{3}$ is $\mathrm{NH}_{4}^{+}$ !

STEP 9

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Studdy Solution

STEP 1

STEP 2

1. Conjugate Bases2. Conjugate Acids

STEP 3

Let's **start** with NH4+\mathrm{NH}_{4}^{+}NH4+​.To find its conjugate base, we **remove** one H+H^{+}H+.So, NH4+−H+=NH3\mathrm{NH}_{4}^{+} - H^{+} = \mathrm{NH}_{3}NH4+​−H+=NH3​.The **conjugate base** of NH4+\mathrm{NH}_{4}^{+}NH4+​ is NH3\mathrm{NH}_{3}NH3​!

STEP 4

STEP 5

Finally, we have HI. **Subtracting** an H+H^{+}H+ gives us HI−H+=I−HI - H^{+} = I^{-}HI−H+=I−.The **conjugate base** of HI is I−I^{-}I−!

STEP 6

STEP 7

STEP 8

Lastly, we have NH3\mathrm{NH}_{3}NH3​. **Adding** an H+H^{+}H+ gives us NH3+H+=NH4+\mathrm{NH}_{3} + H^{+} = \mathrm{NH}_{4}^{+}NH3​+H+=NH4+​.The **conjugate acid** of NH3\mathrm{NH}_{3}NH3​ is NH4+\mathrm{NH}_{4}^{+}NH4+​!

STEP 9

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1. Conjugate Bases
2. Conjugate Acids

Let's start with $\mathrm{NH}_{4}^{+}$ .
To find its conjugate base, we remove one $H^{+}$ .
So, $\mathrm{NH}_{4}^{+} - H^{+} = \mathrm{NH}_{3}$ .
The conjugate base of $\mathrm{NH}_{4}^{+}$ is $\mathrm{NH}_{3}$ !

Finally, we have HI. Subtracting an $H^{+}$ gives us $HI - H^{+} = I^{-}$ .
The conjugate base of HI is $I^{-}$ !

Lastly, we have $\mathrm{NH}_{3}$ . Adding an $H^{+}$ gives us $\mathrm{NH}_{3} + H^{+} = \mathrm{NH}_{4}^{+}$ .
The conjugate acid of $\mathrm{NH}_{3}$ is $\mathrm{NH}_{4}^{+}$ !