The balanced equation for the above reaction is
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2 mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M
NaOH is a strong base so pH will be around 13 to 12. Whatever number of moles of NaOH you have approximately the pH of NaOH will be around 14 13 or 12
Answer:
3.68 grams.
Explanation:
First we <u>convert 9.5 g of NaCl into moles of NaCl</u>, using its<em> molar mass</em>:
9.5 g ÷ 58.44 g/mol = 0.16 mol NaCl
In<em> 0.16 moles of NaCl there are 0.16 moles of sodium </em>as well.
We now <u>convert 0.16 moles of sodium into grams</u>, using <em>sodium's molar mass</em>:
0.16 mol * 23 g/mol = 3.68 g