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86.30 mL of an HCl solution was required to neutralize 31.75 mL of 0.150 M NaOH. Determine the molarity of the HCI.

Sagot :

samueladesida43

Answer:

0.055M

Explanation:

Using the formula as follows:

CaVa = CbVb

Where;

Ca = concentration of acid (M)

Cb = concentration of base (M)

Va = volume of acid (mL)

Vb = volume of base (mL)

According to the information given in this question, Ca (HCl) = ?, Cb (NaOH) = 0.150, Va (HCl) = 86.30mL, Vb (NaOH) = 31.75mL

Using CaVa = CbVb

Ca = CbVb ÷ Va

Ca = (0.150 × 31.75) ÷ 86.30

Ca = 4.7625 ÷ 86.30

Ca = 0.055M

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