Calculate the pOH where [OH⁻]=4.7×10⁻³M: pOH=-log[OH⁻] pOH=-log(4.7×10⁻³) pOH=2.33
Calculate [OH⁻] if the pOH is 1.34: pOH=-log[OH⁻] 10^(-pOH)=[OH⁻] [OH⁻]=10^(-pOH) [OH⁻]=10^(-1.34) [OH-]=0.0457M or [OH⁻]=4.57×10⁻²M
Calculate pH if the [OH⁻] is 1.74×10⁻²M: [H⁺]=K(w)/[OH⁻] or pOH=-log[OH⁻] [H⁺]=(1×10⁻¹⁴)/(1.57×10⁻²) pOH=-log(1.74×10⁻²) [H⁺]=6.369×10⁻¹³M pOH=1.759 pH=-log[H⁺] pH=14-pOH pH=-log(6.369×10⁻¹³) pH=14-1.759 pH=12.2 pH=12.2
I hope this helps. Let me know if anything is unclear.
