Answer:
The method is accurate in the calculation of the 
Explanation:
As a first step we have to calculate the <u>average concentration </u>of find it by the method.
Then we have to find the<u> standard deviation:</u>
For the confidence interval we have to use the formula:
μ=Average±
Where:
t=t student constant with 95 % of confidence and 5 data=2.78
μ= 
± 
upper limit: 0.84
lower limit: 0.75
If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
brainly.com/question/21881762
Answer:
9.51 × 10⁴ kL
Explanation:
Step 1: Given data
Volume of the sample (V): 9.51 × 10⁹ cL
Step 2: Convert "V" to liters
We will use the conversion factor 1 L = 100 cL.
9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L
Step 3: Convert "V" to kL
We will use the conversion factor 1 kL = 1000 L.
9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL
9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.
