Question

Parts 1 & 2, Standardization of NaOH and HC Record the mclarities from standardization or from reagent bottles of the HCl and the NaOH. molarity of HCI1.0780 molarity of NaOH 4851 Your value is acceptable. M Your value is acceptable. M Part 3. Determination of the Amount of Acid Neutralized by an Antacid Tablet Record the data from your titration in Table 1. (Enter NONE in any unused answer blanks.) Table 1 Amount of HCI Initially in Flask (mol) Amount of HCI Neutralized by Titration (mol) Mass of tablet Volume of HCI Volume of NaOH Tablet 25.00 4 25.00 4 25.00 25.00 25.00 425.78 26.19 26.18 926.19 26.09 490 1.48 40 1.46 0 1.49 40 1.50 Averages 1.48 Standard Deviation.02 40 0.02695 0.02695 0.02695 0.02695 9 0.01099 0.25950 0.2595 0.02695 ν' 90.00 0,00 0,00 Determine the number of moles of HCl not neutralized by the NaOH·This is the number of moles of HCl neutralized by the antacid. Record this in Table 2. (Enter NONE in any unused answer blanks.) Table 2 Amount of HCI Neutralized by TabletTablet (mol) E Results Find the average number of moles of HCI neutralized by the tablet and standard deviation. [Enter unrounded values first.

Calculate the Amount of HCl Neutralized by Titration (mol)

Answer 1

Hi,

In a titration, the neutralization is carried out when all the hydroniums of the acids (H +) are neutralized by the hydroxyls of the bases (OH-), at this point the moles of acids are equal to the moles of base, for this reason, the moles of neutralized acids can be known from the added moles of base:

Moles of acid = moles of base
moles of HCl = [NaOH] * Vb

Where [NaOH] is the molar concentration of sodium hydroxide initially 0.4851M and Vb is the volume in liter added of the base sodium hydroxide in each titration.

Molar = M = mol / L

Trial 1
[NaOH] = 0.4851M
Vb = 25.78 ml

Volume of milliliters to liters:


25.78ml1000ml 0.02578/L . 1000m7 = 0.02578し


Moles of Hcl:

moles of HCl = 0.4851-* 0.02578 L

moles of HCI = 0.0125|mol
Trial 2
[NaOH] = 0.4851M
Vb = 26.19 ml

Volume of milliliters to liters:


1L 26.19ml-* 1000777-0.026 1 9L


Moles of Hcl:

moles of HCI 0.485100.02619L

moles of HCl = 0.01270m

Trial 3
[NaOH] = 0.4851M
Vb = 26.18 ml

Volume of milliliters to liters:


26.18ml * -0.02618し 1000ml


Moles of Hcl:

moles of HCl = 0.4851-* 0.02618し

moles of HCI = 001269m이
Trial 4
[NaOH] = 0.4851M
Vb = 26.09 ml

Volume of milliliters to liters:


1L 26.09ml * = 0.02609し 1000ml


Moles of Hcl:
$moles \;of \;HCl = 0.4851\frac{mol}{L}* 0.02609L$
moles of HCI = 001266m


To find the average of the neutralized moles, the moles obtained in each test are added and divided by the number of tests (4):


Moles HCI1+ Moles HCl 2 + Moles HCl 3 +Moles HCl 4 Average moles of HCI
$Average \;moles\; of \;HCl = \frac{0.01251+0.01270+0.01269+0.01266}{4}$

Finally the standard deviation with the help of the excel tool gave a super small value, 8,832 * 10-5, this value to be very small is close to 0.


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