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Calculate the pH of the resultant mixtures: 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

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प्रश्न

Calculate the pH of the resultant mixtures: 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

संख्यात्मक
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उत्तर

Moles of `"H"_3"O"^+`  = `(2xx10xx0.1)/1000` = `.002 " mol"`

Moles of `"OH"^- = (10 xx 0.1)/1000 = 0.001 " mol"`

Excess of `"H"_2"O"^+` = .001 mol

Thus `["H"_3"O"^+] = .001/(20 xx 10^(-3)) =   10^(-3)/(20xx10^(-3))` =  .05

`therefore "pH" = - log (0.05)`

= 1.30

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अध्याय 6: Equilibrium - EXERCISES [पृष्ठ २३८]

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एनसीईआरटी Chemistry - Part 1 and 2 [English] Class 11
अध्याय 6 Equilibrium
EXERCISES | Q 7.66 - c) | पृष्ठ २३८

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