acid base titration calculator
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Answers:Your method is correct, but the exact values I think are not. When doing calculations, save rounding till the end and instead track your sig figs with a line underneath the correct digit. Without rounding during the calculations, you would use 5.125x10^-3 and 1.04475x10^-3. This adds up to a different answer at the end. For the answers I got .400g and 82.6%.
Answers:MW of HCl: 36.463 (1.36 g)(1 mole / 36.463 g) = 0.037298 moles Concentration = (0.037298 moles / 0.500 L) Concentration = 7.4596 x 10^-2 M And the reaction is: HCl ---> H+ + Cl- And since it is assumed that HCl completely disassociates And also since H+ is produced in a 1:1 ratio of HCl that disassociates [H+] = 7.4596 x 10^-2 M pH = - log [H+] pH = 1.13
Answers:write the balanced eq HCl + NaOH --> NaCl + H2O So HCl : NaOH = 1:1 find moles OH- used; n = conc. x volume = 0.2M x 0.025L = 0.005 mol ratio is 1:1 so 0.005 mol HCl neutralised conc. (molarity) = n / volume = 0.005mol / .01833L = 0.27M
Answers:moles HCl = 0.2 x 0.0185 dm^3 =0.0037 moles NaOH = 0.0037 Molarity of NaOH = 0.0037/ 0.025 dm^3=0.148 M