Kemi

Puffer

12. maj 2015 af lalala1122 (Slettet) - Niveau: Universitet/Videregående

Er der nogen der kan hjælpe mig i gang med denne opgave


Brugbart svar (0)

Svar #1
12. maj 2015 af mathon


          pufferligningen        pH=pK_a+\log\left ( \frac{n(NaC_7H_5O_2)}{n(C_7H_6O_2)} \right )

                                          \frac{n(NaC_7H_5O_2)}{n(C_7H_6O_2)} =10^{\, pH-pK_a}

                                          \frac{c(NaC_7H_5O_2)}{c(C_7H_6O_2)} =10^{\, 4{,}00-4{,}20}=0{,}630957

                                         \frac{c(NaC_7H_5O_2)}{c(C_7H_6O_2)+c(NaC_7H_5O_2)} =\frac{0{,}630957}{1+0{,}630957}

                                        \frac{c(NaC_7H_5O_2)}{c_{total}} =\frac{0{,}630957}{1{,}630957}

                                        c(NaC_7H_5O_2) =\frac{0{,}630957}{1{,}630957}\cdot (0{,}100\; M)=0,038686\; \frac{mol}{L}

                                        c(C_7H_6O_2) =\left(0{,}100\; \frac{mol}{L}\right)-(0,038686\; \frac{mol}{L})=0{,}961314\; \frac{mol}{L}
afvejning:
                 m(NaC_7H_5O_2) =\left (0{,}038686\; \frac{mol}{L} \right )\cdot \left ( 144{,}10316\;\frac{ g}{mol} \right )=5{,}575\; \frac{g}{L}

                 m(C_7H_6O_2) =\left (0{,}961314\; \frac{mol}{L} \right )\cdot \left ( 122{,}12134\;\frac{ g}{mol} \right )=117{,}397\; \frac{g}{L}

                                       

                                         


Brugbart svar (0)

Svar #2
13. maj 2015 af mathon

korrektion:

                              c(NaC_7H_5O_2) =\frac{0{,}630957}{1{,}630957}\cdot (0{,}100\; M)=0,038686\; \frac{mol}{L}

                                        c(C_7H_6O_2) =\left(0{,}100\; \frac{mol}{L}\right)-(0,038686\; \frac{mol}{L})=0{,}061314\; \frac{mol}{L}
afvejning:
                 m(NaC_7H_5O_2) =\left (0{,}038686\; \frac{mol}{L} \right )\cdot \left ( 144{,}10316\;\frac{ g}{mol} \right )=5{,}58\; \frac{g}{L}

                 m(C_7H_6O_2) =\left (0{,}061314\; \frac{mol}{L} \right )\cdot \left ( 122{,}12134\;\frac{ g}{mol} \right )=7{,}49\; \frac{g}{L}


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