By definition titraion of a monoprotic acid with means that the equivalence point implies netrality of the solution, which is pH = 7.
So, the answer is that pH will be equal to 7 at the equivalence.
Given that the acid is monoprotic and KOH has one OH- radical per molecule of KOH, the titration will require the same number of moles of acid than base to reach the equivalence point, as you can see in this equation, representing the monoprotic acid as HA:
HA + KOH = K(+) + A(-) + H2O => 1 mol HA per 1 mol KOH.
Answer:
A.All Mixtures Are Made Up Of Solutions
Explanation:
<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_%7B10%7D%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7B%28C_4H_6%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_%7B%28H_2%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-2877.6%29%29%5D-%5B%281%5Ctimes%20%28-2540.2%29%29%2B%282%5Ctimes%20%28-285.8%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D234.2J)
Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.
Answer:
refer the above attachment
<span>Answer:
The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first.
moles HCl = 0.04000 L * 0.100 M = 0.00400 moles
moles KOH = 0.02500 L * 0.100 M = 0.00250 moles
moles HCl left = 0.00400 - 0.00250 = 0.00150 moles
Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+]
pH = -log [H+] = -log (0.0231) = 1.64</span>
