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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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Answer:

the true statement is... The pH of the weak acid will be higher than the pH of the strong acid

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This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

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Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol

The given chemical reaction is:

2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = \frac{3}{2}\times 0.05=0.0750mol of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

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V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = 62.364\text{ L Torr }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

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

743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L

Hence, the volume of hydrogen gas that will be collected is 1.85 L

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