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Artist 52 [7]
2 years ago
15

How should the ph of a 0.1 m solution of nac2h3o2 compare with that of a 0.1 m solution of kc2h3o2?

Chemistry
1 answer:
Semenov [28]2 years ago
7 0

We have that for the Question it can be said that the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

From the question we are told

how should the ph of a 0.1 m solution of <em>nac2h3o2</em> compare with that of a 0.1 m <u>solution </u>of kc2h3o2?

Generally

with  the ph of a 0.1 m solution of <em>nac2h3o2</em> compared with that of a 0.1 m <u>solution </u>of kc2h3o2 ,we see that the salt produce  is a weak acid and strong akali salt

We see that the salt produced in water gives a base from the derived weak acid

The salt produce is CH_3COONa

Therefore

the NaOH combines with CH_3COOH to produce CH_3COONa (Salt)

For more information on this visit

brainly.com/question/17756498

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Which of the following statements best describes the current atomic theory
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In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex
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The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

  NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$.

So at the equivalence point, the moles of NaOH added = moles of $HCl$present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$  mole/L

The volume of NaOH added up to the color change = $V_2 \text{  and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$  moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or   when    $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$  mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$, then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$)

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