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What volume of 0.181 N NaOH is required to neutralize 25.3 mL of 0.114 N H2SO4?
Wiki User
∙ 9y ago
i assume 0.291 and 0.350 are concentration in mol/dm3
----------------------------
NaOH + HCL --> NaCL + H20
no of mol of HCL = 0.350 x (25/1000) = 8.75 x 10^-3
1 mol of HCL react with 1 mol of NaOH
8.75 x 10^-3 mol of HCL react with 8.75 x 10^-3 mol of NaOH
volume = 8.75 x 10^-3/0.291 = 30.1cm3 (3.s.f)
Wiki User
∙ 15y ago
Wiki User
∙ 14y ago0.02612 L * 0.04624 M = 0.0012077888 mole of NaOH in solution.
Since each NaOH donates 1 OH- there is also 0.0012077888 mole of OH-.
This reacts 1 to 1 with the H+, so there is also 0.0012077888 mole of H+
Since each HCl donates 1 H+, there is also 0.0012077888 mole of HCl.
The used HCl solution has 0.1438 mol/liter,
0.0012077888 mole / 0.1438 M = 0.0083990876 liter.
Which is 8.3991 mililiter.
Wiki User
∙ 9y agoAnswer: 15.9 mL NaOH solution
Calculation: When N (Normality, eq/L) is the given unit for concentration (Titre) then simple multiplication-formula is to be used:
Vb * Tb= Va * Ta ,
so
Vb (mL) * 0.181 (eq/L) = 25.3 (mL) * 0.114 (eq/L) and
Vb (mL) = 25.3 * 0.114 / 0.181 (= 15.93) = 15.9 mL NaOH solution.
Wiki User
∙ 6y agoI could really use help with this question and an explanation for my Chemistry homework, Thanks!
Wiki User
∙ 7y agoThe answer is 10 mZL.
Wiki User
∙ 13y ago30 mls
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