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

that's just how they are

explanation:

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Answer from: Quest

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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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