1. **State the problem:** We need to find the critical value for a 95% confidence interval for the population mean nitrate concentration based on a sample of size 28. 2. **Identify the distribution:** Since the population standard deviation is unknown and the sample size is less than 30, we use the t-distribution. 3. **Degrees of freedom:** Calculate degrees of freedom as $df = n - 1 = 28 - 1 = 27$. 4. **Confidence level:** For a 95% confidence interval, the significance level is $\alpha = 1 - 0.95 = 0.05$. 5. **Critical value:** The critical value $t^*$ is the t-score such that the area in the two tails is 0.05, so each tail has 0.025. 6. **Look up or calculate:** Using a t-distribution table or calculator for $df=27$ and $\alpha/2=0.025$, the critical value is approximately $$t^* = 2.052$$. **Final answer:** The critical value to use is $2.052$ (rounded to three decimal places).