Question

A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Part A Determine whether or not 250 mgNaOH would exceed the capacity of the buffer to neutralize it. Determine whether or not 250 would exceed the capacity of the buffer to neutralize it. yes no Request Answer Part B Determine whether or not 350 mgKOH would exceed the capacity of the buffer to neutralize it. Determine whether or not 350 would exceed the capacity of the buffer to neutralize it. yes no Request Answer Part C Determine whether or not 1.25 gHBr would exceed the capacity of the buffer to neutralize it. Determine whether or not 1.25 would exceed the capacity of the buffer to neutralize it. yes no Request Answer Part D Determine whether or not 1.35 gHI would exceed the capacity of the buffer to neutralize it. Determine whether or not 1.35 would exceed the capacity of the buffer to neutralize it. yes no

Answer 1

Answer:

Neither of them will neutralize the buffer solution.

Explanation:

The buffer solution of HNO₂ and KNO₂ will be neutralized when the acid reacts and consume all of the base of the buffer solution or when the base added reacts and consume all of the acid of the buffer solution.

First, we need to calculate the number of moles of the acid and the base of the buffer:

$n_{HNO_{2}} = C*V = 0.100 M*0.500 L = 0.050 moles$

$n_{KNO_{2}} = C*V = 0.150 M*0.500 L = 0.075 moles$

Now, let's evaluate each case.

A) 250 mg of NaOH:

We need to calculate the number of moles of NaOH

$n_{NaOH} = \frac{m}{M}$

Where m: is the mass = 250 mg, and M: is the molar mass = 39.99 g/mol  

$n_{NaOH} = \frac{0.250 g}{39.99 g/mol} = 6.25 \cdot 10^{-3} moles$

The number of moles of the acid HNO₂ after reaction with the base added NaOH is:

$n_{a_{T}} = n_{a} - n_{b} = 0.050 moles - 6.25 \cdot 10^{-3} moles = 0.044 moles$

After the reaction of HNO₂ with the NaOH remains 0.044 moles of acid, hence, 250 mg of NaOH would not exceed the capacity of the buffer to neutralize it.

B) 350 mg KOH:

The number of moles of KOH is:

$n_{KOH} = \frac{m}{M} = \frac{0.350 g}{56.1056 g/mol} = 6.23 \cdot 10^{-3} moles$

Now, the number of moles of HNO₂ that remains in the solution is:

$n_{T} = 0.050 moles - 6.23 \cdot 10^{-3} moles = 0.044 moles$

Therefore, 350 mg of KOH would not exceed the capacity of the buffer to neutralize it.

C) 1.25 g of HBr:

The number of moles of HBr is:

$n_{HBr} = \frac{m}{M} = \frac{1.25 g}{80,9119 g/mol} = 0.015 moles$

Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HBr is:

$n_{T} = 0.075 moles - 0.015 moles = 0.06 moles$

Hence, 1.25 g of HBr would not exceed the capacity of the buffer to neutralize it.

D) 1.35 g of HI:

The number of moles of HI is:

$n_{HI} = \frac{m}{M} = \frac{1.35 g}{127.91 g/mol} = 0.0106 moles$

Now, the number of moles of the base KNO₂ that remains in solution after the reaction with HI is:

$n_{T} = 0.075 moles - 0.0106 moles = 0.0644 moles$

Hence, 1.35 g of HI would not exceed the capacity of the buffer to neutralize it.

Therefore, neither of them will neutralize the buffer solution.

I hope it helps you!