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GaryK [48]
3 years ago
7

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Chemistry
2 answers:
GREYUIT [131]3 years ago
7 0

Answer:

(a) before addition of any KOH  = pH =-log [0.210]=0.68

(b) after addition of 25.0 mL of KOH  = pH =-log [0.210]=1.15

(c) after addition of 35.0 mL of KOH  = pH =-log [0.210]=1.43

(d) after addition of 50.0 mL of KOH  = pH =-log [0.210]=7.00

(e) after addition of 60.0 mL of KOH = pH =-log [0.210]=12.30

Explanation:

pH is calculated as follows:

pH = -log [H+], for acids

pOH = -log [OH-], for bases

pH = 14 - pOH

In this tritation, we add a base (KOH), to an acid (HClO), which neutralices each other in a stoiquiometric relationship of 1:1

So, reemplacing the values in each case we got:

(a) before addition of any KOH

: The concentration is equal to the concentration of HClO, beacuse no KOH has been added

<em>pH =-log [0.210]=0.68</em>

(b) after addition of 25.0 mL of KOH  

The concentration is equal to the amount of moles of HClO, that haven´t reacted with the moles of KOH cantained in the 25 ml added, divided by the total volumen (50 ml + 25 ml) = 75 ml

to find the amount of moles in the 25ml, and the 50 ml:

moles KOH = volumen L * Molarity = 0.025 ml * 0.210 M = 5.25*10^{-3}

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*10^{-2}

moles HClO - moles KOH = 1.05*10^{-2}-5.25*10^{-3}=5.25*10^{-3}

concentration = moles / volume = \frac{5.25*10^{-3}moles of HClO}{0.075ml} = 0.070M

<em>pH =-log [0.070]=1.15</em>

(c) after addition of 35.0 mL of KOH  

The concentration is equal to the amount of moles of HClO, that haven´t reacted with the moles of KOH cantained in the 35 ml added, divided by the total volumen (50 ml + 35 ml) = 85 ml

to find the amount of moles in the 35ml, and the 50 ml:

moles KOH = volumen L * Molarity = 0.035 ml * 0.210 M = 7.35*10^{-3}

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*10^{-2}

moles HClO - moles KOH = 1.05*10^{-2}-7.35*10^{-3}=3.15*10^{-3}

concentration = moles / volume = \frac{3.15*10^{-3}moles of HCl}{0.085ml} = 0.037M

<em>pH =-log [0.037]=1.43</em>

(d) after addition of 50.0 mL of KOH  

moles HCl = moles KOH

pH = 7

(e) after addition of 60.0 mL of KOH

The concentration is equal to the amount of moles of KOH, that haven´t reacted with the moles of HClO cantained in the initial 50 ml, because now the KOH is in excess, divided by the total volumen (50 ml + 60 ml) = 100 ml

to find the amount of moles in the 50ml, and the 60 ml:

moles KOH = volumen L * Molarity = 0.060 ml * 0.210 M = 1.26*10^{-2}

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*10^{-2}

moles KOH - moles HClO = 1.26*10^{-2}-1.05*10^{-2}=2.1*10^{-3}

concentration = moles / volume = \frac{2.1*10^{-3}moles of KOH}{0.110ml} = 0.020M

<em>pOH =-log [0.020]=1.70</em>

<em>pH = 14 - pOH = 12.3</em>

laiz [17]3 years ago
3 0
<span>35.0 mL of 0.210 M
KOH molarity = moles/volume
 find moles of OH do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+ they will cancel out: H+ + OH- -> H2O
 but you'll have some left over,
 pH=-log[H+] pOH
     =-log[OH-] pH+pOH
     =14</span>
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