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What volume of 0.300 M NaOH solution is required to neutralize 32.5 cm of a
0.180 M HCl solution?

What Volume Of 0300 M NaOH Solution Is Required To Neutralize 325 Cm Of A 0180 M HCl Solution class=

Sagot :

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Explanation:

NaOH (aq) + HCl(aq) ----> NaCl(aq) + H2O(l)

Stoichiometric ratio NaOH : HCl = 1:1

moles of HCl needed to be neutralized = 0.18/1000 × 32.5 = 5.85 × 10^-3 mol

So 5.85 × 10^-3 mol of NaOH should be used

If that amount is found in V volume

5.85 × 10^-3 = 0.3/1000 ×V

5.85/0.3 cm^3 = 19.5cm^3

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