The molarity of the HCl solution needed to neutralize 28.6 mL of a 0.175 M NaOH solution is 0.2002 M
We'll begin by writing the balanced equation for the reaction. This is given below:
HCl + NaOH —> NaCl + H₂O
From the balanced equation above,
The mole ratio of the acid, HCl (nA) = 1
The mole ratio of the base, NaOH (nB) = 1
- From the question given above, the following data were obtained:
Volume of base, NaOH (Vb) = 28.6 mL
Molarity of base, NaOH (Mb) = 0.175 M
Volume of acid, HCl (Va) = 25 mL
Molarity of acid, HCl (Ma) = ?
The molarity of the acid, HCl can be obtained as follow:
MaVa / MbVb = nA / nB
(Ma × 25) / (0.175 × 28.6) = 1
(Ma × 25) / 5.005 = 1
Cross multiply
Ma × 25 = 5.005
Divide both side by 25
Ma = 5.005 / 25
Ma = 0.2002 M
Therefore, the molarity of the acid, HCl needed for the reaction is 0.2002 M
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