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miskamm [114]
3 years ago
10

charcoal removes fine particulates, as well as lead, mercury, copper, and chlorine from water through a separation process calle

d
Chemistry
1 answer:
Pachacha [2.7K]3 years ago
7 0
Absorption distillation floatation filtration
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Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to
Mama L [17]

We have the given reaction as;

HCl_{(aq)} + NaOH_{(aq)} ----> NaCl_{(aq)} + H_{2}O_{(l)}<span>

Answer A) The pH will be 12.36,</span>

<span>We have to convert the concentrations of HCl and NaOH into moles,</span>

So we have, n(HCl) = (0.0290 L) X (0.290 mol/L) = 8.41X 10^{-3} moles <span>

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X 10^{-2} moles 

Now, it seems NaOH is in excess, so amount remaining will be; 

1.13 X 10^{-2} - 8.41 X 10^{-3} = 2.89 X 10^{-3} moles 

<span>Now, the total volume will become as  = 0.0390 + 0.0290 = 0.068 L </span>

So, the concentration of [OH^{-}] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X 10^{-2} M 

pOH = - log [OH^{-}] = -log (4.25×10^{-2}) = 1.37 

Hence, pH = 14 - pOH = 14 - 1.37 = 12.6</span>

So the pH of the solution will be 12.6 which is basic in nature. <span>

Answer B) The pH will be 1.68 </span>

<span>Now, for the given concentration we need to find moles for HCl and NaOH also;</span>

n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X 10^{-3} mol <span>

<span>n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X  10^{-3} mol </span>

here we can see, HCl is in excess amount so the remaining will be;  

8.41X 10^{-3} - 7.41 X 10^{-3} = 1.0 X 10^{-3} mol 

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L 

So the concentration of [HCl] = 1.0 X 10^{-3} mol / 0.0480 L = 2.08 X 10^{-2} M </span>

 

Which is = [H⁺] <span>

So, the pH = - log [H^{+}] = -log(2.08X 10^{-2}) = 1.68</span>

 

Hence, the pH will be 1.63 which is more acidic in nature.


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3 years ago
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