How would you solve this problem? ?

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.

668900 atoms in total

Particle exchange energy though elastic collision , hope this .