For a liter of a buffer that is 1. 5 m in acetic acid and 0. 70 m in sodium acetate, and with a result in a buffer pH of 4.01, the moles of HCl required is mathematically given as x=0.313mol
<h3>What is the result in a buffer ph of 4. 01?</h3>
Generally, the equation for the pH is mathematically given as
pH=pKa+logsalt/acid
Generally, the equation for the Chemical Reaction is mathematically given as
HCl+CH3CooNa---->NaCl+CH3OOH
Therefore
4.01=4.74+log\frac{0.8-x}{1.3+x}
-0.52=log\frac{0.8-x}{1.3+x}
x=0.313mol
In conclusion, The moles of HCl required is
x=0.313mol
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The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 0.125 mL
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the volume of HCl neutralized is 0.125 mL
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Its electronic confuguration is 2.8.6 so it occupies 3 shells.