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Advocard [28]
4 years ago
14

What is the ph of a solution that has 0.075 m ch3cooh and 0.050m nach3coo present? ka for acetic acid = 1.80 x 10-5?

Chemistry
1 answer:
Scilla [17]4 years ago
7 0
Hello!

For solving this problem we first have to determine the pKa of CH₃COOH.

pKa=-log(Ka)=-log( 1,80*10^{-5} )=4,74

Now, we use the Henderson-Hasselbach's equation to determine the pH of the buffer solution:

pH=pKa+log (\frac{NaCH_3COO}{CH_3COOH} )=4,74+log( \frac{0,050M}{0,075M} )=4,56

So, the pH of this solution is 4,56

Have a nice day!
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The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.
mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

6 0
4 years ago
What mass of fes is formed if 9.42 g of fe reacts with 68.0 g of s8? 8fe(s) + s8(s) → 8fes(s)?
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From the equation, we see that the molar ratio of Fe : S required is:

8 : 1

The moles of Fe present are: 9.42/56 = 0.168
Moles of S = 68/(32 * 8) = 0.265

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1 : 1.6

Therefore, iron is the limiting reactant as it is present in a ratio lower than that required. The ratio of

Fe : FeS  is
1 : 1

So 0.168 moles of FeS will form. The mass of FeS will be:

Mass = 0.168 * (56 + 32)
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