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UNO [17]
3 years ago
6

1. What is the relationship between the concentration of the hydronium and hydroxide

Chemistry
1 answer:
Hitman42 [59]3 years ago
4 0

Answer:

- Neutral solutions: concentration of hydronium equals the concentration of hydroxide.

- Acid solutions: concentration of hydronium is greater than the concentration of hydroxide.

- Basic solutions concentration of hydronium is lower than the concentration of hydroxide.

Explanation:

Hello,

It is widely known that the pH of water is 7, therefore the pOH of water is also 7 based on:

pH+pOH=14

In such a way, we can compute the concentration of hydronium and hydroxide ions as shown below:

pH=-log([H^+])\\

[H^+]=10^{-pH}=10^{-7}=1x10^{-7}M

pOH=-log([OH^-])

[OH^-]=10^{-pOH}=10^{-7}=1x10^{-7}M

Thus, we notice that the relationship between the concentration of the hydronium is equal for water or neutral solutions. Moreover, if we talk about acid solutions, pH<OH therefore the concentration of hydronium is greater than the concentration of hydroxide. On the other hand if we talk about basic solutions, pH>OH therefore the concentration of hydronium is lower than the concentration of hydroxide.

Best regards.

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A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is
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3 years ago
You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

Now put all the given values in the above formula, we get:

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

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3 years ago
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