<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer:
96g
Explanation:
Given parameters:
Number of moles of NaOH = 2.4moles
Unknown:
Mass of NaOH = ?
Solution:
The mass of a substance given the number of moles can be found using the expression below;
Mass of NaOH = Number of moles x molar mass;
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Now input the parameters and solve;
Mass of NaOH = 2.4 x 40 = 96g
Answer:
Is this chemistry or ELA!
V=0,1 l=100 ml
m=p*V=0,8787*100=87,87g
