The pH of a solution of HN3 (Ka=1.9×10−5) and NaN3 is 4.86.I figured this out but according to my HW it is wrong so can someone show me how to do this? pH=Pka+log [N3-]/[HN3] Pka=-log (ka) so Pka= -log(1.9*10^-5)= 4.7212log [N3-]/[0.018]= 4.86-4.72 log [N3-]/[0.018]= 0.1387 10^(0.014)=1.387[N3-]/0.018=1.387[N3-]=0.02MWhat am I doing wrong?. Created by Courtney. science-mathematics-en - chemistry-en

you ask wrong q:pH = pKa -lg(acid/base)4,86 = 4,72 -lg(0,018/base)lg(...) = -0,140,018/base = 0,72[base] = 0,025 M

What is the pH of a solution of HN3 (Ka=1.9×10−5) and NaN3 is 4.86?