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Sagot :
Answer:
Following are the responses to the given choices:
Explanation:
For point a:
Using the acid and base which are strong so,
moles of [tex]H^+[/tex] (from[tex]HNO_3[/tex])
[tex]= 24.9\ mL \times 0.100\ M \\\\= \frac{24.9}{1000\ L} \times 0.100\ M \\\\= 2.49 \times 10^{-3} \ mol[/tex]
moles of [tex]OH^{-}[/tex] (from [tex]KOH[/tex])
[tex]= 25.0\ mL \times 0.100\ M \\\\= \frac{25.0}{1000 \ L} \times 0.100 \ M \\\\\= 2.50 \times 10^{-3}\ mol[/tex]
[tex]1\ mol H^{+} \ neutralizes\ 1\ mol\ of\ OH^{-}[/tex]
So, [tex](2.50 \times 10^{-3} mol - 2.49 \times 10^{-3} mol)[/tex] i.e. [tex]1 \times 10^{-5}[/tex] mol of [tex]OH^-[/tex] in excess in total volume [tex](24.9+25.0) \ mL = 49.9 \ mL[/tex] i.e. concentration of [tex]OH^- = 2 \times 10^{-4}\ M[/tex]
[tex]p[OH^{-}] = -\log [OH^{-}] = -\log [2 \times 10^{-4}\ mol] = 3.70[/tex]
Since, [tex]pH + pOH = 14,[/tex]
so,
[tex]\to pH = 14- pOH = 14- 3.70 = 10.30[/tex]
For point b:
moles of [tex]OH^-[/tex] = from point a [tex]= 2.50 \times 10^{-3} \ mol[/tex]
moles of [tex]H^+[/tex](from[tex]HNO_3[/tex]):
[tex]= 25.1 mL \times 0.100 M\\\\ = \frac{25.1}{1000}\ L \times 0.100 \ M\\\\ = 2.51\times 10^{-3} \ mol[/tex]
1 mol [tex]H^+[/tex] neutralizes 1 mol of [tex]OH^-[/tex]
So, [tex](2.51 \times 10^{-3}\ mol - 2.50 \times 10^{-3}\ mol)[/tex] i.e. [tex]1 \times 10^{-5} \ mol \ of\ H^+[/tex] in excess in the total volume of [tex](25.1+25.0) \ mL = 50.1\ mL[/tex] i.e. concentration of[tex]H^+ = 2 \times 10^{-4}\ M[/tex]
Hence, [tex]pH = -\log [H^+] = -\log[2 \times 10^{-4}] = 3.70[/tex]
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