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s2008m [1.1K]
3 years ago
9

A chemist needs to create a series of standard Cu2+(aq) solutions for an absorbance experiment. For the first standard, he uses

a pipet to transfer 10.00 mL of a 2.61 M Cu2+(aq) stock solution to a 250.0 mL volumetric flask, and he adds enough water to dilute to the mark. He then uses a second pipet to transfer 10.00 mL of the second solution to a 100.0 mL volumetric flask, and he adds enough water to dilute to the mark. Calculate the concentration of the Cu2+(aq) solution in the 100.0 mL volumetric flask. concentration: M
Chemistry
1 answer:
morpeh [17]3 years ago
6 0

Answer:

0.01 M

Explanation:

The chemist is performing a serial dilution in order tyo obtain the calibration curve for the instrument.

First we must obtain the concentration of the solution in the 250ml flask from

C1V1 = C2V2

Where;

C1 = concentration of the stock solution

V1 = volume of the stock solution

C2 = concentration of the diluted solution

V2= volume of the diluted solution

2.61 × 10 = C2 × 250

C2 = 2.61 × 10/250

C2 = 0.1 M

Hence for solution in 100ml flask;

0.1 × 10 = C2 × 100

C2 = 0.1 × 10/100

C2 = 0.01 M

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of
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Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

mol=0.282*0.01

mol=0.00282

We can do the same for the KOH values (50mL=0.05L and 0.141M).

0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

0.00705-0.00282=0.00423mol~of~KOH

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Total~volume=50mL+10mL=60mL

60mL=0.06L

Now we can calculate the concentration:

M=\frac{0.00423mol}{0.06L}

M=0.0705

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

pOH=-Log(0.0705)

pOH=1.15

Now we can <u>calculate the pH value</u>:

14=~pH~+~pOH

pH=14-1.15=12.84

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