Question

Part A

Before you begin, you collect and determine the following information:

Molarity of acid 0.141 M
Volume of acid (mL) 25.00 mL
Volume of acid (L) 0.025 L
Imagine your flask contains HCl (hydrochloric acid) and your burette contains KOH (potassium hydroxide). You can add the base in large or small increments. You stop when the color inside the flask is a faint pink color. Remember, you want the color to be a faint pink color, not dark pink. If you notice the color changing but then returning to colorless, that tells you that you are getting close to the equivalence point. You continue titrating until the faint pink color remains. And you enter your data in the table below:

Volume of base (ml) 35.38 mL
Volume of base (l) 0.03538 L

Write a balanced equation for this reaction.


Part B
From your data above, fill in the following values:
MOLARITY OF ACID
VOLUME OF ACID
MOLARITY BASE................CURRENTLY UNKNOWN
VOLUME OF BASE

Part C
Calculate the number of moles of HCl that were in your flask.

Part D
How many moles of NaOH will there be?

Part E
Calculate the concentration of the NaOH in the titration.

Answer 1

A) HCl + NaOH → NaCl + H2O

B) MOLARITY OF ACID:0.141 M

VOLUME OF ACID:25mL

Volume of base: 35.38 mL

C) volume (L) × concentration (M) = moles HCl

0.025 L times; 0.141 M = 0.0035 moles

D) 0.00353 moles NaOH

E) moles of NaOH ÷ volume of NaOH(L) = molarity of NaOH

0.00353 moles NaOH ÷ 0.03528 = 0.100 M NaOH

Step-by-step explanation:

edmentum

Answer 2

Part A : The balanced chemical equation is,

`HCl(aq)+KOH(aq)\rightarrow KCl(aq)+H_2O(aq)`

Part B : Molarity of Acid = 0.141 mole/L

Volume of Acid = 25 ml = 0.025 L

Molarity of Base = 0.0996 mole/L

Volume of Base = 35.38 ml = 0.03538 L

Part C : The number of moles of HCl = 0.003525 moles

Part D : The number of moles of NaOH = 0.003523 moles

Part E : The concentration of NaOH = 0.0996 mole/L

Solution : Given,

Molarity of Acid = 0.141 mole/L

Volume of Acid = 25 ml = 0.025 L

Volume of Base = 35.38 ml = 0.03538 L

Part A : The balanced chemical equation is,

`HCl(aq)+KOH(aq)\rightarrow KCl(aq)+H_2O(aq)`

Part B : Acid = HCl Base = NaOH

Molarity of Acid = 0.141 mole/L

Volume of Acid = 25 ml = 0.025 L

Volume of Base = 35.38 ml = 0.03538 L

Now we have to calculate the Molarity of NaOH.

Formula used :
`M_(HCl)* V_(HCl)=M_(NaOH)* V_(NaOH)`

where,

`M_(HCl)` = Molarity of HCl

`M_(NaOH)` = Molarity of NaOH

`V_(HCl)` = Volume of HCl

`V_(NaOH)` = Volume of NaOH

Now put all the given values in this formula, we get the value of
`M_(NaOH)`.

`0.141mole/L* 0.025L=M_(NaOH)* 0.03538L\\M_(NaOH)=0.0996mole/L`

Therefore, The Molarity of Base (NaOH) = 0.0996 mole/L

Part C : we have to calculate the number of moles of HCl.

Formula used :
`n_(HCl)=M_(HCl)* V_(HCl)`

where,

`n_(HCl)` = Moles of HCl

`M_(HCl)` = Molarity of HCl

`V_(HCl)` = Volume of HCl

Now put the given values in above formula, we get

`n_(HCl)=0.141mole/L* 0.025L=0.003525moles`

Therefore, the number of moles of HCl = 0.003525 moles

Part D : we have to calculate the number of moles of NaOH.

Formula used :
`n_(NaOH)=M_(NaOH)* V_(NaOH)`

where,

`n_(NaOH)` = Moles of NaOH

`M_(NaOH)` = Molarity of NaOH

`V_(NaOH)` = Volume of NaOH

Now put the given values in above formula, we get

`n_(NaOH)=0.0996mole/L* 0.03538L=0.003523moles`

Therefore, the number of moles of NaOH = 0.003523 moles

Part E : we have to calculate the concentration of NaOH.

Formula used :
`C_(NaOH)=(n_(NaOH))/(V_(NaOH))`

where,

`C_(NaOH)` = Concentration of NaOH

`n_(NaOH)` = Moles of NaOH

`V_(NaOH)` = Volume of NaOH

Now put the given values in above formula, we get

`C_(NaOH)=(0.003523moles)/(0.03538L)=0.0996mole/L`

Therefore, the concentration of NaOH = 0.0996 mole/L