V ( H2SO4) = 35 mL / 1000 => 0.035 L
M ( H2SO4) = ?
V ( NaOH ) = 25 mL / 1000 => 0.025 L
M ( NaOH ) = 0.320 M
number of moles NaOH:
n = M x V
n = 0.025 x 0.320 => 0.008 moles of NaOH
Mole ratio:
<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??
0.008 x 1 / 2 => 0.004 moles of H2SO4 :
Therefore:
M ( H2SO4) = n / V
M = 0.004 / 0.035
= 0.114 M
hope this helps!
Answer:
1. pH = 1.23.
2. 
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the pKa is:

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

Which is also shown in net ionic notation.
Best regards!
We have to add the both half cell equations and eliminate the number of electrons lost/gained.
<h3>What modification must Kim make to the equations?</h3>
The term redox reaction is a type of reaction that occurs when an electron is lost or gained in a reaction system. We can see that in this reaction, zinc looses two electron which are gained by copper.
If we want to obtain the equation 4.9 which is the overall equation of the redox reaction from the various half cell equations then we have to add the both half cell equations and eliminate the number of electrons lost/gained.
Learn kore about redox reaction:brainly.com/question/13293425
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