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3 years ago

CH3CN major species present when dissolved in water

2 answers:

5 0

The major species present when CH3CN is dissolved in water are CH3+ and CN-. It doesn't completely dissociate in water since it is not a strong acid or strong base. Only a strong acid or base will dissociate completely in water. For example HCl dissociates into H+ and Cl-, and NaOH dissociates into Na+ and OH-.

Margaret [11]3 years ago

3 0

Answer: CH₃CN and H₂O.

Explanation:

1) The spieces present in a solution may be either the molecules, in case of covalent compounds, or ions, in case of ionic compounds that dissociate (ionize).

2) Both, CH₃CN and H₂O are covalent (polar covalent) substances, so they do not ionize and the spieces in the solution are the molecules per se.

3) In solution, the molecules of H₂O will solvate the molecules of CH₃CN, meaning that H₂O molecules are able to separate the molecules of CH₃N from each other, and so every molecule of CH₃CN will end surrounded by many molecules of H₂O.

This happens because the interaction between the polar molecules of the two different compounds is strong enough to overcome the intermolecular forces between the molecules of the same compound.

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3 years ago

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3 years ago

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8 0

3 years ago

Read 2 more answers

Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to

Mama L [17]

We have the given reaction as;

$HCl_{(aq)} + NaOH_{(aq)}$ ----> $NaCl_{(aq)} + H_{2}O_{(l)}$ 

Answer A) The pH will be 12.36,

We have to convert the concentrations of HCl and NaOH into moles,

So we have, n(HCl) = (0.0290 L) X (0.290 mol/L) = 8.41X $10^{-3}$ moles 

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X $10^{-2}$ moles

Now, it seems NaOH is in excess, so amount remaining will be;

1.13 X $10^{-2}$ - 8.41 X $10^{-3}$ = 2.89 X $10^{-3}$ moles

Now, the total volume will become as = 0.0390 + 0.0290 = 0.068 L 

So, the concentration of [ $OH^{-}$ ] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X $10^{-2}$ M

pOH = - log [ $OH^{-}$ ] = -log (4.25× $10^{-2}$ ) = 1.37

Hence, pH = 14 - pOH = 14 - 1.37 = 12.6

So the pH of the solution will be 12.6 which is basic in nature. 

Answer B) The pH will be 1.68

Now, for the given concentration we need to find moles for HCl and NaOH also;

n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X $10^{-3}$ mol 

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X $10^{-3}$ mol 

here we can see, HCl is in excess amount so the remaining will be;

8.41X $10^{-3}$ - 7.41 X $10^{-3}$ = 1.0 X $10^{-3}$ mol

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L

So the concentration of [HCl] = 1.0 X $10^{-3}$ mol / 0.0480 L = 2.08 X $10^{-2}$ M

Which is = [H⁺] 

So, the pH = - log [ $H^{+}$ ] = -log(2.08X $10^{-2}$ ) = 1.68

Hence, the pH will be 1.63 which is more acidic in nature.

6 0

3 years ago

Read 2 more answers