Don't add the CIs (confidence intervals, by the way, not concentration intervals) of the slope and intercept, because the slope and intercept are strongly correlated, so you can't add their variances as if they were independent. That's why you get such wide limits. More detailed analysis of the regression gives (IIRC) the following formula for the confidence interval:

CI = ±t_{95%}*s/sqrt(n)*{1 + (x - x)^{2}/(x^{2} - x^{2})}^{1/2}

where t is Student's t-statistic, s is the standard error, and I have written means with underlines because I can't work out how to write overlines. (The means refer to the set of n data points used to calculate the regression line.) Note how this increases as x deviates from x.

Actually this refers to the confidence interval of the regression value of y for a particular value of x. More relevant for you is the prediction interval, which is the likely deviation of a single measurement from the regression line. This is given by

PI = ±t_{95%}*s/sqrt(n)*{n + 1 + (x - x)^{2}/(x^{2} - x^{2})}^{1/2}

However, all this refers to the deviation of a y measurement for a given x value. What you want is the deviation of the estimate of x from a measured y value. There is a formula for this, and I worked it out myself once long ago, but I was told by a real analytical chemist not to use it, because it gives a spurious precision to your result - actually, given all the other possible sources of error in a measurement, the uncertainty in your result is much greater than the regression confidence interval. So unfortunately I can't really answer your question!