Chemistry, 16.03.2020 23:01 thanks5640

A buffer solution contains 0.353 M ammonium bromide and 0.352 M ammonia. If 0.0200 moles of hydrochloric acid are added to 125 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume change does not change upon adding hydrochloric acid)

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Answer from: gonzalesalexiaouv1bg

9.07

Explanation:

We have to start with the buffer system reaction, so:

NH_4^+~~NH_3~+~H^+

When we have the hydrochloric acid (a strong acid) the H^+ of the hydrochloric acid ( HCl ) will interact with the base of the buffer system ( NH_3 ) to produce more acid ( NH_4^+ ), so:

HCl~+~NH_3-~NH_4^+~+Cl^-

Therefore the concentration of NH_3 will decrease and the concentration of NH_4 will increase. The next step then would be the calculation of the moles of the acid and base in the buffer system. So:

$M=\frac{#mol}{L}$

#mol=0.353*0.125=0.044~mol~of~NH_4^+

#mol=0.352*0.125=0.044~mol~of~NH_3

If we add 0.02 mol of HCl we can calculate the amount of acid and base that changes in the buffer system.

0.044~mol~of~NH_3~-~0.02=0.024

0.044~mol~of~NH_4^+~+~0.02=0.064

Now, we can calculate the concentration of each species if we divide by the number of moles:

$M=\frac{0.024~mol}{0.0125~L}=~0.192~M~of~NH_3$

$M=\frac{0.064~mol}{0.0125~L}=~0.512~M~of~NH_4^+$

If we use the hendersson hasselbach equation we can calculate the pH value again:

$pH=p{ K }_{ a }+log(\frac { { [A }^{ - }] }{ [HA] } )$

$pH=9.5+log(\frac {0.192}{0.512} )$

pH=9.07

The final pH value is 9.07

Answer from: Quest

Answer earth,water,fire and air;

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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