0.1 mole of CH₃NH₂ (k_b = 5 x 10^-4 ) is added to 0.08 moles of HCl and the solution is diluted to 1 litre , then calculate resulting hydrogen ion concentration in the solution .

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Solution

CH3NH2 + HCl = CH3NH3+ + Cl-
Initial 0.1 0.08 -
At Eq. 0.02 - 0.08
[OH-] = Kb [CH3NH2]/[CH3NH3+]
= 510^-4 0.02/0.08
= 1.25 * 10^-4
[H+] = Kw/[OH-]
= 10^-14 * 4/ 510^-4
= 8 10^-11

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$0.1 m o l$ of methyl amine ( $K_{b}$ = $5 \times 10^{- 4}$ ) is mixed with $0.08 m o l$ of $H C l$ and the solution is diluted to $1 l i t r e$ . The hydrogen ion concentration of the resulting solution is:

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0.1 mole of CH3NH2 (kb = 5 x 10-4 ) is added to 0.08 moles of HCl and the solution is diluted to 1 litre , then calculate resulting hydrogen ion concentration in the solution .

CH3NH2 + HCl = CH3NH3+ + Cl-Initial 0.1 0.08 -At Eq. 0.02 - 0.08[OH-] = Kb [CH3NH2]/[CH3NH3+]= 5*10^-4 * 0.02/0.08= 1.25 * 10^-4[H+] = Kw/[OH-]= 10^-14 * 4/ 5*10^-4= 8 * 10^-11

0.1 mole of CH₃NH₂ (k_b = 5 x 10^-4 ) is added to 0.08 moles of HCl and the solution is diluted to 1 litre , then calculate resulting hydrogen ion concentration in the solution .

CH3NH2 + HCl = CH3NH3+ + Cl-
Initial 0.1 0.08 -
At Eq. 0.02 - 0.08
[OH-] = Kb [CH3NH2]/[CH3NH3+]
= 510^-4 0.02/0.08
= 1.25 * 10^-4
[H+] = Kw/[OH-]
= 10^-14 * 4/ 510^-4
= 8 10^-11