In the titration of 20.00 mL of 0.175 M NaOH, calculate the number of milliliters of 0.200 M HCl that must be added to reach a pH of 12,55.
I know the answer is 11,9 mL but I don't know how to calculate it.
NaOH + HCl ==> NaCl + H2O
You want pH = 12.55 which means the solution will be quite basic so all of the NaOH has not been neutralized.
14-12.55 = pOH = 1.45 and (OH^-) = -log(OH^-) so (OH^-) = 0.0355M.
If we let x = mL 0.2M HCl added, then mmols OH^- remaining in the titration will be 0.0355*(20.00 + x).
We start with 20.00*0.175 = 3.500 mmols NaOH. As we titrate it the NaOH will be
(3.500-0.2x). Set these equal to each other to obtain
(3.500-0.2x) = 0.0355(20+x) and solve for x = mL 0.2M HCl. Watch the significant figures (I didn't) and check it when you are finished. I worked it out and obtained 11.85 mL which when plugged back into the problem gives a pH of 12.55. I would round that 11.85 to 11.8 but your prof may not round like I do. You may be able to tweak it here and there in the equation but I believe this is the way to work it.
Well, well, well, looks like we've got ourselves a math problem! Don't worry, I'm here to help you out, with a touch of clown humor, of course.
To solve this problem, we can use the concept of stoichiometry. In a neutralization reaction between an acid and a base, the moles of acid react with an equal number of moles of base. So, let's find the number of moles of NaOH first.
Moles of NaOH = concentration × volume
Moles of NaOH = 0.175 M × 0.020 L
Moles of NaOH = 0.0035 mol
Since the acid and base react in a 1:1 ratio, we need the same number of moles of HCl to neutralize the NaOH. Now, let's find the volume of HCl needed.
Volume of HCl = Moles of HCl ÷ Concentration
Volume of HCl = 0.0035 mol ÷ 0.200 M
Volume of HCl ≈ 0.0175 L
Finally, to convert L to mL, we multiply by 1000.
Volume of HCl = 0.0175 L × 1000 mL/L
Volume of HCl ≈ 17.5 mL
Oops, it seems like I've made a clown mistake here! My apologies for the confusion. The correct volume of 0.200 M HCl needed to reach a pH of 12.55 would actually be around 17.5 mL, not 11.9 mL. So, grab your clown glasses and measure out that HCl!
To solve this problem, we need to use the concept of stoichiometry and the balanced equation for the reaction between NaOH and HCl, which is:
NaOH + HCl -> NaCl + H2O
We are given the volume and concentration of NaOH (20.00 mL, 0.175 M) and the desired pH (12.55). We want to determine the volume of HCl (in mL) that needs to be added to reach this pH.
Step 1: Write the balanced equation for the reaction
NaOH + HCl -> NaCl + H2O
Step 2: Calculate the number of moles of NaOH used
moles of NaOH = volume of NaOH (L) x concentration of NaOH (M)
moles of NaOH = (20.00 mL / 1000 mL) x 0.175 M
moles of NaOH = 0.0035 moles
Step 3: Determine the number of moles of HCl required
Since NaOH and HCl react in a 1:1 ratio, the moles of HCl required will be equal to the moles of NaOH used.
moles of HCl required = moles of NaOH used = 0.0035 moles
Step 4: Calculate the volume of HCl required
volume of HCl (L) = moles of HCl required / concentration of HCl (M)
volume of HCl (L) = 0.0035 moles / 0.200 M
volume of HCl (L) = 0.0175 L
Step 5: Convert the volume from L to mL
volume of HCl (mL) = 0.0175 L x 1000 mL/L
volume of HCl (mL) = 17.5 mL
Therefore, the number of milliliters of 0.200 M HCl that must be added to reach a pH of 12.55 is approximately 17.5 mL, which differs from your given answer of 11.9 mL. Please check your calculations or provide more information if necessary.
To calculate the number of milliliters of 0.200 M HCl needed to reach a pH of 12.55 in the titration of 20.00 mL of 0.175 M NaOH, you need to understand the concept of neutralization and stoichiometry.
The reaction between NaOH (sodium hydroxide) and HCl (hydrochloric acid) is a neutralization reaction, where sodium chloride (NaCl) and water (H2O) are formed.
1. Write the balanced chemical equation for the reaction:
NaOH + HCl -> NaCl + H2O
From the balanced equation, we can see that the ratio between NaOH and HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.
2. Calculate the number of moles of NaOH:
moles NaOH = volume (in L) x concentration (in mol/L)
= 0.020 L x 0.175 mol/L
= 0.0035 moles of NaOH
3. Since the ratio between NaOH and HCl is 1:1, the number of moles of HCl needed is also 0.0035 moles.
4. Calculate the volume of HCl using its concentration:
volume HCl (in L) = moles HCl / concentration HCl (in mol/L)
= 0.0035 moles / 0.200 mol/L
= 0.0175 L
5. Finally, convert the volume from liters to milliliters:
volume HCl (in mL) = volume HCl (in L) x 1000 mL/L
= 0.0175 L x 1000 mL/L
= 17.5 mL
Therefore, you would need to add approximately 17.5 mL of 0.200 M HCl to reach a pH of 12.55, which is different from your expected answer of 11.9 mL. Double-check the problem or provide more information if you believe there is an error.