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For $pu550.0 mL$ that a buffer equipment that is $pu0.170 M$ in $ceCH3CH2NH2$ and $pu0.150 M$ in $ceCH3CH2NH3Cl$, calculation the early pH and also the last pH after adding $pu0.020 mol$ that $ceHCl$.

The adhering to is my effort on fixing this problem:

Since Kb because that ethylalamine = $5.6 imes 10^-4$.

pKb = $-log( 5.6 imes 10^-4) = 3.25$

pOH = $3.25 + log (0.150/0.170)$

pOH = $3.196$

pH = $14 - 3.196 = 10.80$ because that Initial

Final pH:

$pu0.170 M * pu0.55 L= 0.0935 mol$ $ceCH3CH2NH2$

$pu0.150 M * pu0.55 L = pu0.0825 mol ceCH3CH2NH3Cl$,

$0.0825 - 0.02 = pu0.0625 mol ceCH3CH2NH3Cl$,

$0.0935 + 0.02 = pu0.1135 mol ceCH3CH2NH2$

pOH = $3.25 + log(0.0625/0.1135)$

pOH = $2.991$

pH = $14 - 2.991 = 11.01$ for final

However, this values revolve out to it is in wrong. You re welcome help, say thanks to you!

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edited Mar 30 "19 at 20:58

Tyberius
asked Mar 30 "19 in ~ 3:22

toffeetoffee
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Part A- Calculating the early $pupH$(before adding $pu0.02 mol$ of $ceHCl)$

The equilibrium here is $$ceCH3CH2NH2 + H2O CH3CH2NH3+ + OH−$$For $ce CH3CH2NH2 , mathrmK_b = 4.6 imes10^−4$

$$mathrmK_b = 4.6 imes10^−4 =frac$$The nominal concentrations of the base and also conjugate acid space respectively $$mathrmC_b =pu0.170 M~ ext and mathrm C_a =pu0.150 M$$.

See more: What Is The Hybridization Of Carbon In Co32−? What Is The Hybridization Of Carbon In Co3 2

The massive balance expression are:$$+ = mathrmC_a + mathrmC_b = pu0.32M~ ext and~ = mathrmC_a=pu0.150M$$and the charge balance is provided by:$$+ = + approx$$in i m sorry the approximation shown over is justified the the systems will be weakly alkaline.

The equilibrium concentration of the conjugate types are then:$$= 0.150 +$$

eginarray &= &(mathrmC_a + mathrmC_b) &-&()\ &= &(0.32) &–&()\ &= &0.320 &- &(0.150 + )\&=&0.170 &- &endarrayBecause $mathrmC_a$ and $mathrmC_b$ are large compared come $$(the equipment is not intended to be strong alkaline),the$$ state in the over expressions can be dropped and the equilibrium expression becomes$$mathrmK_b=5.6 imes10^-4 =frac (0.150)(0.170)$$from i m sorry we uncover $= pu6.347 imes10^-4 M$ , $pupOH = 3.2$ and $pupH = 14- 3.2 = 10.8$

Part B : Calculating the last $pupH$(after including $pu0.020 mol$ the $ce HCl$)

Calculate the initial amount of $ce CH3CH2NH3+$ and also the initial quantity of $ceCH3CH2NH2$:eginalignn_ce CH3CH2NH3+ &=n_ce CH3CH2NH3Cl\ &=V_ ext(solution) imes_ extI\ &=fracpu550.0 mlpu1000 ml/L imespu0.150 mol/L\&=pu0.083 mol endaligneginalignn_ce CH3CH2NH2 &=V_ ext(solution) imes_mathrmI\ &=fracpu550 ml pu1000 ml/L imespu0.170 mol/L\&=pu0.094 mol endalignAddition that the solid acid $pu 0.020 mol$ of $ce HCl$ , spend this same amount that $ce CH3CH2NH2$ and produces $pu 0.020 mol$ of $ce CH3CH2NH3+$ ion as described in the complying with table:

eginarray c c c c c c c extrxn&ce CH3CH2NH2 &+ &ceHCl &-> &ceCH3CH2NH3+ &+ &ceCl-\hline extI&(0.094)& &(0.020)& &(0.083)& &(0.083)\ extC &(0.094-0.020)& &(0.020-0.020)& &(0.083+ 0.020)& &(0.083+ 0.020) \ extF &(pu0.074 mol)& &(pu0.000 mol)& &(pu0.103 mol)& &(pu0.103 mol) endarray

We can continue using the approximations because that the concentration of the conjugate speciesthat we developed in the preceding part A.The values of C$_mathrmb$ and C$_mathrma$ space now

$$mathrmC_mathrmb = _mathrmI =fracn_ceCH3CH2NH2V=fracpu0.074molpu0.550 L=pu0.133 Mapprox_mathrme$$and$$mathrmC_mathrma = _ extI =fracn_ceCH3CH2NH3+V=fracpu0.103 molpu0.55 L=pu0.187 Mapprox_mathrme$$

Substituting these right into the equilibrium constant expression yieldseginalign &=(mathrmK_b)frac \&=(5.6 imes10^-4) imesfracpu0.133 M pu0.187M\&=pu3.983 imes10^-4 M endalignthe brand-new $pupOH= 3.40$ ,and the new $pupH=14- 3.40=10.60$, so addition of the acid has changed the $pupH$, a decrease of just $0.20~~pupH$ unit.