Question

Answer this question fast, I'll give u 50 points. 100 mL of 0.1 M KOH is mixed with 200 mL of 0.1 M Ca(OH )2 and 300 mL of 0.1 M H2SO4 . Thus, resultant mixtures are

Answer 1

The mixture is acidic which has 0.017N

Given:

100mL of 0.1M KOH is mixed with 200mL of 0.1M Ca(OH)2 and 300mL of 0.1M H2SO4

To Find:

The resultant of mixtures obtained by the two

Solution:

It is a case of mixture of acid and base. Compare N1V1(acid) with N2V2 (base)

First determine the normality of mixture of bases,

0.1M KOH = 0.1N

KOH = 0.1N [O$H^{-}$]

0.1M Ca(OH)2 = 0.2N,

Ca(OH)2 = 0.2N [O$H^{-}$]

Resultant normality of base,

N(base) = ΣNV/ΣV

= $\frac{N1V1(KOH)+N2V2(Ca(oH)2}{V1+V2}$

=$\frac{0.1 (100) +0.2(200)}{100+200}$

= $\frac{N}{6}$

It is mixed with 300mL of 0.1 M h2so4 = 300mL of 0.2 N$H^{+}$

Mixture of acidic,

N$H^{+}$ = $\frac{N1V1-N2V2}{V1+V2}$

= $\frac{60-50}{600}$

=0.017

Hence, the resultant of mixture is acidic which has the value 0.017

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