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:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
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:
![n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.01M\\V_2=25mL](https://tex.z-dn.net/?f=n_1%3D1%5C%5CM_1%3D2M%5C%5CV_1%3D%3FmL%5C%5Cn_2%3D1%5C%5CM_2%3D0.01M%5C%5CV_2%3D25mL)
Putting values in above equation, we get:
![1\times 2\times V_1=1\times 0.01\times 25\\\\V_1=\frac{1\times 0.01\times 25}{1\times 2}=0.125mL](https://tex.z-dn.net/?f=1%5Ctimes%202%5Ctimes%20V_1%3D1%5Ctimes%200.01%5Ctimes%2025%5C%5C%5C%5CV_1%3D%5Cfrac%7B1%5Ctimes%200.01%5Ctimes%2025%7D%7B1%5Ctimes%202%7D%3D0.125mL)
Hence, the volume of HCl neutralized is 0.125 mL
Answer:
Option b=> 0.1 M CH3COOH , 0.1 M CH3COONa.
Explanation:
In chemistry, when an acid or a base(alkali) is added to a solution and the solution pH does not change then we say that the solution is a BUFFER SOLUTION (that is a solution, upon the addition of acid or base will show resistance in pH value).
In order to be able to answer this question efficiently we have to consider the mathematical representation or Equation below;
pH = pKa + log ( [ A^-] / [HA] )--------------------------------------------------------------------(1).
What the mathematical representation or Equation (1) above is telling is is that if [A^-] or [HA] is high then there will be the production of buffer with higher buffer capacity.
Answer:
Explanation:
to show how precise your measurements were.
<span>A 18 M solution of an acid that ionizes only slightly in solution would be termed
concentrated and weak. The concentration of the acid is high. The acid which dissociates partially in water is a weak acid.
</span><span>Calculate the [H^+] for the aqueous solution in which [OH^-] is 1 x10^-9. Is this solution acidic, basic or neutral. To determine [H+] use:
1x10^-14 = [OH-][H+]
solve for [H+]
[H+] = 1x10^-14/1x10^-9
= 1x10^-5</span>
Answer:
269.068 kJ/mol.
Explanation:
<em>ln (k₂/k₁) = (Eₐ/R) [(T₂ - T₁)/(T₁T₂)].</em>
<em>k₁ = 6.20 x 10⁻⁴ min⁻¹, T₁ = 700.0 K.</em>
<u><em>To get k₂:</em></u>
in first order reactions: k = 0.693/(half-life).
∴ k₂ = 0.693/(29.0 min) = 2.39 x 10⁻² min⁻¹, T₂ = 760.0 K.
∵ ln (k₂/k₁) = (Eₐ/R) [(T₂ - T₁)/(T₁T₂)]
∴ ln [(2.39 x 10⁻² min⁻¹)/(6.20 x 10⁻⁴ min⁻¹)] = (Eₐ/(8.314 J/mol.K)) [(760.0 K - 700.0 K) / (760.0 K)(700.0 K)].
3.65 = (Eₐ/(8.314 J/mol.K)) (1.128 x 10⁻⁴).
<em>∴ Eₐ =</em> (3.65)(8.314 J/mol.K) / (1.128 x 10⁻⁴) = <em>269.068 kJ/mol.</em>