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.
