Ka = [H=][ClO2-]/[HClO2] = 0.011
Let [H+] = x. Then [ClO2-] = x and [HClO2] = 0.155 - x. But x is very small, so we can neglect it and say that [HClO2] = 0.155.
Ka = (x)(x)/(0.155) = 0.011
x^2 = 0.01705
x = 0.1306 [H+]
pH = -Log[H+], and Log(0.1306) = -0.8841, so pH = +0.8841
