25 ml of a solution of Na2CO3 having a specific gravity of 1.25 gram ml required 32.9 ml of a solution of HCl containing 109.5 gram of the acid per litre for complete neutralization. Calculate the volume of 0.84 N H2SO4 that will be completely neutralized by 125 gram of Na2CO3 solution.

Since Na2CO3  is completely neutralized by HCl Also, given molarity of HCl = 109.536.5=3Therefore, milliequivalents of Na2CO3 = milliequivalents of HCl N×25 = 32.9×3 N = 3.948Now, a fresh Na2CO3 solution reacts with H2SO4 Volume of 125 gm of  Na2CO3 = 1251.25 = 100 mlAgain we can write,Milliequivalents of  Na2CO3 = Milliequivalents of H2SO4 100×3.948 = 0.84×V V =470 ml