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elena55 [62]
3 years ago
14

A buffer solution is made that is 0.417 M in and 0.417 M in . If for is , what is the pH of the buffer solution? pH = Write the

net ionic equation for the reaction that occurs when 0.120 mol is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter.)
Chemistry
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

pH = 3.347

NO₂⁻ + H₃O⁺ → HNO₂

Explanation:

If the buffer is made with 0.417M of HNO₂ and 0.417M of NaNO₂ the pH of the solution is = pKa. <em>A buffer which concentration of weak acid and conjugate base is the same has a pH = pKa</em>

As Ka of the solution is 4.50x10⁻⁴, pKa is -log 4.50x10⁻⁴ = 3.347. Thus,<em> pH = 3.347</em>

<em />

The net ionic equation that occurs when 0.120 mol of HCl is added to 1.00L of the solution is:

<em>NO₂⁻ + H₃O⁺ → HNO₂ + H₂O</em>

<em>The conjugate base (NaNO₂) reacts with strong acid (HCl, source of H₃O⁺) producing weak acid (HNO₂) and water. If a strong base is added, the weak acid reacts producing conjugate base and water.</em>

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1%

Explanation:

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7 0
3 years ago
In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
11 months ago
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3 years ago
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Answer:

Explanation:

Here are the ones I'm certain of.

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9. Contains just one sigma bond. H2 is made up of only 1 bond type. I'd pick this, but It's not certain.

=============

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