[Chemical analyst] Titrate 50mL of 0.02M weak acid (pKa=6.72) solution into 0.1MNaOH solution. What is the pH at the equivalent point?
Add 50mL of a 0.02M weak acid (pKa=6.72) solution to 0.1MNaOH solution. What is the pH at the equivalent point? ① Neutralization reaction by NaOH A Hydrolysis reaction of A-anions HA+OH→A-+H2O(acid base neutralization reaction)$NaOH solution volume required =\frac{0.02M\\times50mL}{0.10M}=10mL$NaOH solution volume required = 0.02M×50mL0.10M=10mL $\left[\combi{]A}^-\right] =\frac{0.02M\x50mL}{\left(50+10\right)mL}\ = 0.01667M\ $[A-] = = 0.02M×50mL(50+10)mL = = 0.01667M [A-][HA]-[OH-]I0.0166700E0.01667-××$\combi{pK}_b=14-\combi{pK}_a=14-6。72=7.28ドルpKb=14-pKa=14-6。72=7.28$\combi{K}_b\ […]