I need to see these questions solved. For potentiometric detections, you can alw

Question

I need to see these questions solved. For potentiometric detections, you can always use the flowing equation to calculate the amount of analtyte 0.05916 V E . constant + loglanalytel -- n Answer questions 1 and 2 based on this equation. Assume you are measuring pH value of your sample using a glass electrode. You first immersed it in a pH 4.01 standard solution and the potential reading was 0.511V. After you immerse the electrode in your sample solution, a voltage of 0.523 was displayed. What is the pH of your sample? 1. 2. Assume you are using a So selective electrode to detect a sample. When you put the electrode in a standard S solution (o.01M), a potential of 0.322 V was recorded. When you put the same electrode in your sample, a potential of 0.411 V was displayed. What's the molarity of S in the sample?

Explanation / Answer

I will answer question 1 and 2. The third question post it in another question thread.

1. pH is a value that can be determined by the following equation:

pH = -log[H+]

so in this case, we can assume that n = 1 cause we are measuring a solution with some concentration of H+ transfering one proton. With the first data, we will calculate the constant. And then, with this value and the other data we will calculate the pH

Rearranging the equation with the value of pH we have:

E = K - (0.05916/n) pH

Let's solve for K:

K = E + (0.05916/n)pH

K = 0.511 + 0.05916*4.01 = 0.7482 V

Now, solving for pH:

E - K = -0.05916pH

-(E - K)/0.05916 = pH

pH = 0.7402 - 0.523 / 0.05916

pH = 3.67

For question 2, you do it in a similar way, but instead of pH we calculate the concentration at the end:

K = E - (0.05916/n)logC

K = 0.322 - 0.05916/2 log 0.01 = 0.38116 V

Now for concentration:

E - K = 0.05916/2 logC

2(E-K) / 0.05916 = logC

logC = 2(0.411 - 0.38116) / 0.05916

logC = 0.50439

C = 3.1944 M

Hope this helps

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---