The equation of the reaction in mention is;
#H_2SO_4 + 2NaOH -> Na_2SO_4 + 2H_2O#
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Using this formula to get the concentration of the base,
#(C_aV_a)/(C_bV_b) = (n_a)/(n_b)#
Where;
#C_a = "concentration of the acid"#
#V_a = "volume of the acid"#
#C_b = "concentration of the base"#
#V_b = "volume of the base"#
#n_a/n_b = "mole ratio of the acid to the base"#
#C_a = 0.05 M#
#V_a = 40 mL#
#C_b = ?M#
#V_b = 50mL#
#n_a/n_b = 1/2#
Please note that #n_a and n_b# are gotten based on the stoichiometric coefficients (mole ratios) from the equation.
Hence substituting the values into the equation.
#(0.05 xx 40)/(Cb xx 50) = 1/2#
#2/(50C_b) = 1/2#
Cross multiplying,
#1 xx 50C_b = 2 xx 2#
#50C_b = 4#
Dividing both sides by #50#
#(50C_b)/50 = 4/50#
#(cancel50C_b)/cancel50 = 4/50#
#C_b = 4/50#
#C_b = 0.08M or 0.08molL^-1#
Now to calculate the Normality of the #NaOH# solution.
Note: Normality is the moles equivalent of the concentration per litre.
#"Normality" = "moles equivalence"/"litre"#
Since 1 mol of NaOH = 1 equivalent..
Therefore, the normal concentration (normality) is 0.08 eq/L
#"Normality" = 0.08/1#
#"Normality" = 0.08 N#
Hope this helps!
