Chemistry, 07.05.2020 00:12 nick607

A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate the pHpH after the addition of 13.0 mLmL of HNO3

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Answer from: priscillaoliver5214

The pH is pH = 9.4

Explanation:

From the question we are told that

The volume of NH_3 is $V_N = 75mL = 75 *10^{-3} L$

The concentration of NH_3 is C_N = 0.200M

The concentration of HNO_3 is C_H = 0.500 M

The volume of HNO_3 added is $V_H = 13mL = 13 *10^{-3 } L$

The base dissociation constant is $K_b = 1.8*10^{-5}$

The number of moles of HNO_3 that was titrated can be mathematically represented as

$n__{H}} = C_H * V_H$

substituting values

$n__{H}} = 0.500* 13*10^{-3}$

$n__{H}} = 0.0065 \ moles$

The number of moles of NH_3 that was titrated can be mathematically represented as

$n__{N}} = C_N * V_N$

substituting values

$n__{N}} = 0.200 * 75*10^{-3}$

$n__{N}} = 0.015 \ mole$

So from the calculation above the limited reactant is HNO_3

The chemical equation for this reaction is

$NH_3 + HNO_3 ------ NH^{4+} + NO^{3+}$

From the chemical reaction

1 mole of HNO_3 is titrated with 1 mole of NH_3 to produce 1 mole of NH^{4+}

0.0065 moles of HNO_3 is titrated with 0.0065 mole of NH_3 to produce 0.0065 mole of $NH^{4+}$

The remaining moles of NH_3 after the titration is

$n = n__{N}} - n__{H}}$

=> n = 0.015 - 0.0065

$n = 0.0085 \ moles$

Now according to Henderson-Hasselbalch equation the pH of the reaction is mathematically represented as

$pH = pK_a + log [\frac{NH_3}{NH^{4+}} ]$

Where pK_b is mathematically represented as

pK_a = -log K_a

Now $K_a = \frac{K_w}{K_b}$

Where K_w is the ionization constant of NH_3 with value $K_w = 1.0*10^{-14}$

Hence $K_a = \frac{1.0*10^{-14}}{1.8 *10^{-5}}$

$K_a = 5.556 * 10^{-10}$

Substituting this into the equation

$pH = -log K_a + log [\frac{NH_3}{NH^{4+}} ]$

$pH = log [\frac{\frac{NH_3}{NH^{4+}} }{K_a} ]$

substituting values

$pH = log [\frac{\frac{0.0085}{0.0065} }{5.556*10^{-10}} ]$

pH = 9.4

Answer from: Quest

acid rain does not usually kill trees directly. instead, it is more likely to weaken the trees by damaging their leaves, limiting the nutrients available to them, or poisoning them with toxic substances slowly released from the soil. the main atmospheric pollutants that affect trees are nitrates and sulphates.

Answer from: Quest

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

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A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. C...

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