A sample of seawater collected at a beach in Singapore was analysed. It was found to have a turbidityof 15 NTU. For reference, the range of turbidity of drinking water in Singapore is 0.05 – 0.51 NTU.Raw sample Treated sampleTurbidity (NTU) 15 1Absorbance(unitless)0.7 0.05(a) At 25°C, the pKw of seawater is 13.22. This value is lower than that for water as a result of thedissolved salts. If the pH of a sample of seawater is 6.78, what is the pOH of the sample? (1 point)(b) At pH of 6.78 and pKw of 13.22, is this sample of seawater acidic, basic or neutral? Briefly explainwhy. (2 points)(c) The sample is treated to reduce its turbidity. Name one method that can be used to reduce theturbidity of the water sample. (1 point)(d) What is the % transmittance of the raw sample? (1 point)(e) A spectrometer is used to analyse the absorbance of the treated sample. Determine the % of lightabsorbed in the analysis of the treated sample (assume negligible reflection and scattering).(2 points)

(a) The pOH of a sample can be calculated using the formula pOH = pKw - pH. Substituting the given values, we get pOH = 13.22 - 6.78 = 6.44.

(b) A solution is considered acidic if the pH is less than 7, basic if the pH is greater than 7, and neutral if the pH is exactly 7. Since the pH of the seawater sample is 6.78, which is less than 7, the seawater is acidic.

(c) There are several methods to reduce the turbidity of water. One common method is sedimentation, where the water is allowed to sit undisturbed while heavier particles sink to the bottom due to gravity. The clear water can then be decanted or siphoned off.

(d) The % transmittance of a sample can be calculated using the formula %T = 10^(-Absorbance) * 100. Substituting the given absorbance value of the raw sample, we get %T = 10^(-0.7) * 100 = 19.95%.

(e) The % of light absorbed can be calculated using the formula %A = (1 - 10^(-Absorbance)) * 100. Substituting the given absorbance value of the treated sample, we get %A = (1 - 10^(-0.05)) * 100 = 1.12%.